Complex Numbers

The sum can be expanded as follows: \[ S = i^1 + i^2 + i^3 + i^4 + i^5 + \dots + i^{2026} \] Notice that the sum of any four consecutive powers of $i$ is exactly zero: \[ i^{4k+1} + i^{4k+2} + i^{4k+3} + i^{4k+4} = i - 1 - i + 1 = 0 \] To evaluate $S$, we divide the total number of terms ($2026$) by $4$: \[ 2026 = 4 \times 506 + 2 \] This means there are $506$ full cycles of four that sum to zero, leaving exactly the last two terms remaining: \[ S = i^{2025} + i^{2026} \] Using the cyclic properties of $i$: \[ i^{2025} = i^{4(506)+1} = i^1 = i \] \[ i^{2026} = i^{4(506)+2} = i^2 = -1 \] Thus, the sum simplifies to $i - 1$. Final Answer: $-1 + i$.

First, convert each negative radicand into expressions using the imaginary unit $i$ to avoid arithmetic errors: \[ \sqrt{-9} = i\sqrt{9} = 3i \] \[ \sqrt{-16} = i\sqrt{16} = 4i \] \[ \sqrt{-25} = i\sqrt{25} = 5i \] Now substitute these values back into the expression: \[ E = (3i) \times (4i) + 5i \] Calculate the product term: \[ (3i) \times (4i) = 12i^2 = 12(-1) = -12 \] Combine the real and imaginary components: \[ E = -12 + 5i \] Final Answer: $-12 + 5i$.

For an integral power of $i$ to equal $1$, the exponent must be a multiple of $4$. Let us solve for when the quadratic exponent satisfies this condition: \[ x^2 - 7x + 12 = 4k, \quad \text{where } k \in \mathbb{Z} \] The problem states that $x$ must be a real number ($x \in \mathbb{R}$). Let us rearrange the equation into a standard quadratic form: \[ x^2 - 7x + (12 - 4k) = 0 \] For this quadratic equation to yield real roots, its discriminant must be non-negative ($D \ge 0$): \[ D = (-7)^2 - 4(1)(12 - 4k) \ge 0 \] \[ 49 - 48 + 16k \ge 0 \implies 1 + 16k \ge 0 \implies k \ge -\frac{1}{16} \] Since $k$ must be an integer, the values it can take are $k \in \{0, 1, 2, 3, \dots\}$. Let us check the simplest baseline value where $k = 0$: \[ x^2 - 7x + 12 = 0 \implies (x-3)(x-4) = 0 \implies x = 3 \text{ or } x = 4 \] Both $x=3$ and $x=4$ are valid real numbers that satisfy the equation. Final Answer: $x = 3, 4$ (for $k=0$).

Expand the left-hand side of the equation: \[ (x + iy)(2 - 3i) = 2x - 3ix + 2iy - 3i^2y \] Since $i^2 = -1$, group the terms into real and imaginary parts: \[ (2x + 3y) + i(2y - 3x) = 4 + i \] Equate the real and imaginary components from both sides to form a system of simultaneous linear equations: \[ 2x + 3y = 4 \quad \text{--- (Equation 1)} \] \[ -3x + 2y = 1 \quad \text{--- (Equation 2)} \] Let us solve this system using elimination. Multiply Equation 1 by $3$ and Equation 2 by $2$: \[ 6x + 9y = 12 \] \[ -6x + 4y = 2 \] Add the two equations together to eliminate $x$: \[ 13y = 14 \implies y = \frac{14}{13} \] Substitute this value for $y$ back into Equation 1 to solve for $x$: \[ 2x + 3\left(\frac{14}{13}\right) = 4 \implies 2x + \frac{42}{13} = 4 \implies 2x = 4 - \frac{42}{13} = \frac{10}{13} \implies x = \frac{5}{13} \] Final Answer: $x = \frac{5}{13}, y = \frac{14}{13}$.

To eliminate the imaginary unit from the denominator, multiply both the numerator and the denominator by the complex conjugate of the denominator ($1 + 2i$): \[ z = \frac{1 + 3i}{1 - 2i} \times \frac{1 + 2i}{1 + 2i} \] Expand the expressions: \[ \text{Numerator} = (1 + 3i)(1 + 2i) = 1 + 2i + 3i + 6i^2 = 1 + 5i - 6 = -5 + 5i \] \[ \text{Denominator} = (1 - 2i)(1 + 2i) = 1^2 - (2i)^2 = 1 - 4(-1) = 1 + 4 = 5 \] Combine the simplified numerator and denominator: \[ z = \frac{-5 + 5i}{5} = \frac{-5}{5} + i\frac{5}{5} = -1 + i \] The real part is $\text{Re}(z) = -1$ and the imaginary part is $\text{Im}(z) = 1$. Final Answer: $-1 + i$.

Substitute $z = x + iy$ into the rational expression: \[ w = \frac{2(x+iy)+1}{i(x+iy)+1} = \frac{(2x+1) + 2iy}{(1-y) + ix} \] Multiply both the top and the bottom by the conjugate of the denominator, which is $(1-y) - ix$: \[ w = \frac{[(2x+1) + 2iy][(1-y) - ix]}{(1-y)^2 + x^2} \] Expand the numerator to find the imaginary part ($\text{Im}(w)$): \[ \text{Im}(\text{Numerator}) = -(2x+1)x + 2y(1-y) = -2x^2 - x + 2y - 2y^2 \] Set the imaginary part of the full expression equal to $-2$: \[ \frac{-2x^2 - 2y^2 - x + 2y}{x^2 + (1-y)^2} = -2 \] Cross-multiply and simplify: \[ -2x^2 - 2y^2 - x + 2y = -2(x^2 + 1 - 2y + y^2) \] \[ -2x^2 - 2y^2 - x + 2y = -2x^2 - 2y^2 + 4y - 2 \] Cancel out the quadratic terms $-2x^2$ and $-2y^2$ from both sides: \[ -x + 2y = 4y - 2 \implies x + 2y - 2 = 0 \] This is the equation of a straight line. Final Answer: $x + 2y - 2 = 0$.

Let $w = \frac{z-i}{z+i}$. The problem states that $w$ is a purely imaginary number. By definition, a number is purely imaginary if it satisfies the condition $w + \bar{w} = 0$: \[ \frac{z-i}{z+i} + \overline{\left(\frac{z-i}{z+i}\right)} = 0 \] Apply the conjugate distribution property over quotients and sums ($\overline{z_1/z_2} = \bar{z}_1/\bar{z}_2$): \[ \frac{z-i}{z+i} + \frac{\bar{z}-\bar{i}}{\bar{z}+\bar{i}} = 0 \] Since $\bar{i} = -i$, rewrite the expression: \[ \frac{z-i}{z+i} + \frac{\bar{z}+i}{\bar{z}-i} = 0 \] Combine the terms over a common denominator: \[ \frac{(z-i)(\bar{z}-i) + (\bar{z}+i)(z+i)}{(z+i)(\bar{z}-i)} = 0 \] Expand the numerator expressions: \[ (z\bar{z} - iz - i\bar{z} + i^2) + (z\bar{z} + i\bar{z} + iz + i^2) = 0 \] Group like terms and cancel out opposite terms ($-iz$ cancels with $+iz$, and $-i\bar{z}$ cancels with $+i\bar{z}$): \[ 2z\bar{z} + 2i^2 = 0 \] Since $i^2 = -1$, divide the equation by $2$: \[ z\bar{z} - 1 = 0 \implies z\bar{z} = 1 \] Using the property $z\bar{z} = |z|^2$, we get $|z|^2 = 1 \implies |z| = 1$. This is the equation of a unit circle centered at the origin. Final Answer: Proved ($|z| = 1$).

Let $z = x + iy$. Then its conjugate is $\bar{z} = x - iy$. Substitute these variables into the equation: \[ 2(x + iy) + 3(x - iy) = 5 - i \] Expand and group the real and imaginary parts: \[ (2x + 2iy) + (3x - 3iy) = 5 - i \implies (2x + 3x) + i(2y - 3y) = 5 - i \] \[ 5x - iy = 5 - i \] Equate the real and imaginary components from both sides: \[ 5x = 5 \implies x = 1 \] \[ -y = -1 \implies y = 1 \] Reconstruct the complex number $z = x + iy$: \[ z = 1 + i \] Final Answer: $1 + i$.

We can apply the conjugate power distribution property $\overline{(z^n)} = (\bar{z})^n$ to make the calculation simpler: \[ \overline{(1+i)^4} = (\overline{1+i})^4 = (1-i)^4 \] Now evaluate $(1-i)^2$ first: \[ (1-i)^2 = 1^2 - 2i + i^2 = 1 - 2i - 1 = -2i \] Raise this result to the second power to find the fourth power: \[ (1-i)^4 = [(-2i)^2] = 4i^2 = 4(-1) = -4 \] Final Answer: $-4$.

To find the upper bound for $|z|$, use the lower bound branch of the Triangle Inequality: \[ |z_1 + z_2| \ge ||z_1| - |z_2|| \] Let us rewrite $|z|$ to include the expression given in the problem statement: \[ |z| = \left| \left(z - \frac{4}{z}\right) + \frac{4}{z} \right| \] Apply the upper bound branch of the Triangle Inequality ($|A+B| \le |A| + |B|$): \[ |z| \le \left| z - \frac{4}{z} \right| + \left| \frac{4}{z} \right| \] Substitute the given value $|z - \frac{4}{z}| = 2$ and use the modulus quotient property ($|\frac{4}{z}| = \frac{4}{|z|}$): \[ |z| \le 2 + \frac{4}{|z|} \] Multiply the entire inequality by the positive value $|z|$ to clear the denominator: \[ |z|^2 \le 2|z| + 4 \implies |z|^2 - 2|z| - 4 \le 0 \] Solve the corresponding quadratic boundary equation using the quadratic formula: \[ |z| = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-4)}}{2(1)} = \frac{2 \pm \sqrt{20}}{2} = 1 \pm \sqrt{5} \] Since a modulus must be non-negative, the valid interval for the inequality is: \[ 0 \le |z| \le \sqrt{5} + 1 \] Thus, the maximum possible value that $|z|$ can reach is $\sqrt{5} + 1$. Final Answer: $\sqrt{5} + 1$.

• $|2+i| = \sqrt{2^2 + 1^2} = \sqrt{5}$
• $|3-2i| = \sqrt{3^2 + (-2)^2} = \sqrt{9 + 4} = \sqrt{13}$
• $|1+2i| = \sqrt{1^2 + 2^2} = \sqrt{5}$
• $|2-i| = \sqrt{2^2 + (-1)^2} = \sqrt{5}$

According to the upper bound of the Triangle Inequality: \[ |z_1 + z_2| \le |z_1| + |z_2| \] Substitute the given values: \[ |z_1 + z_2| \le 3 + 4 = 7 \] Thus, the maximum value is $7$. This maximum value is reached when the two complex vectors are perfectly aligned along the same ray from the origin, which means their arguments must be equal: $\arg(z_1) = \arg(z_2)$. Final Answer: $7$, when $\arg(z_1) = \arg(z_2)$.

Step 1: Calculate the modulus radius $r$: \[ r = |z| = \sqrt{(-1)^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = \sqrt{4} = 2 \] Step 2: Identify the geometric quadrant of the point. The coordinates are $(-\text{Real}, +\text{Imaginary}) = (-1, \sqrt{3})$, which lies in the Second Quadrant.
Step 3: Calculate the baseline reference angle $\alpha$ using the positive tangent ratio: \[ \tan \alpha = \left| \frac{\text{Im}(z)}{\text{Re}(z)} \right| = \left| \frac{\sqrt{3}}{-1} \right| = \sqrt{3} \implies \alpha = \frac{\pi}{3} \] Step 4: Use the quadrant rule to find the argument $\theta$. For the second quadrant, $\theta = \pi - \alpha$: \[ \theta = \pi - \frac{\pi}{3} = \frac{2\pi}{3} \] Step 5: Assemble the components into standard polar form $r(\cos\theta + i\sin\theta)$: \[ z = 2\left(\cos\frac{2\pi}{3} + i\sin\frac{2\pi}{3}\right) \] Final Answer: $2\left(\cos\frac{2\pi}{3} + i\sin\frac{2\pi}{3}\right)$.

Write out the polar expression and distribute the radius value: \[ z = 4\left(\cos\frac{7\pi}{6} + i\sin\frac{7\pi}{6}\right) \] Evaluate the trigonometric values for the angle $\frac{7\pi}{6}$ (which lies in the third quadrant): \[ \cos\frac{7\pi}{6} = \cos\left(\pi + \frac{\pi}{6}\right) = -\cos\frac{\pi}{6} = -\frac{\sqrt{3}}{2} \] \[ \sin\frac{7\pi}{6} = \sin\left(\pi + \frac{\pi}{6}\right) = -\sin\frac{\pi}{6} = -\frac{1}{2} \] Substitute these values back into the distributed expression: \[ z = 4\left(-\frac{\sqrt{3}}{2} + i\left(-\frac{1}{2}\right)\right) = -2\sqrt{3} - 2i \] Final Answer: $-2\sqrt{3} - 2i$.

• $z_1 = 0 + 2i \implies$ This maps to the point $(0, 2)$, which sits directly on the Positive Imaginary Axis. The ray from the origin to this point points straight up, forming a $90^\circ$ angle with the positive real axis. Thus, $\arg(z_1) = \frac{\pi}{2}$.
• $z_2 = -3 + 0i \implies$ This maps to the point $(-3, 0)$, which sits directly on the Negative Real Axis. The ray points straight left, forming a $180^\circ$ angle. Thus, $\arg(z_2) = \pi$.

Step 1: Find the baseline reference angle $\alpha$ using the absolute values of the components: \[ \alpha = \arctan\left| \frac{\text{Im}(z)}{\text{Re}(z)} \right| = \arctan\left| \frac{-1}{-1} \right| = \arctan(1) = \frac{\pi}{4} \] Step 2: Identify the geometric quadrant. The components are $(-\text{Real}, -\text{Imaginary})$, so the point lies in the Third Quadrant.
Step 3: Apply the principal argument rule for the third quadrant ($\theta = -\pi + \alpha$) to ensure the angle falls within the allowed range $(-\pi, \pi]$: \[ \theta = -\pi + \frac{\pi}{4} = -\frac{3\pi}{4} \] Notice that using the alternative positive angle layout gives $\frac{5\pi}{4}$, which is incorrect because it is greater than $\pi$. Final Answer: $-\frac{3\pi}{4}$.

We can find the argument of a fraction quickly by using the argument distribution property, which converts division into subtraction: $\arg\left(\frac{z_1}{z_2}\right) = \arg(z_1) - \arg(z_2)$.
Calculate the principal argument of the numerator ($z_1 = 1 + i\sqrt{3}$): It lies in the first quadrant. $\alpha_1 = \arctan(\sqrt{3}/1) = \frac{\pi}{3} \implies \arg(z_1) = \frac{\pi}{3}$.
Calculate the principal argument of the denominator ($z_2 = 1 + i$): It also lies in the first quadrant. $\alpha_2 = \arctan(1/1) = \frac{\pi}{4} \implies \arg(z_2) = \frac{\pi}{4}$.
Now subtract the two arguments: \[ \theta = \arg(z_1) - \arg(z_2) = \frac{\pi}{3} - \frac{\pi}{4} = \frac{4\pi - 3\pi}{12} = \frac{\pi}{12} \] Since $\frac{\pi}{12}$ lies within the interval $(-\pi, \pi]$, it is the valid principal argument. Final Answer: $\frac{\pi}{12}$.

Use trigonometric half-angle identities to simplify the real and imaginary parts of the expression: \[ 1 - \cos\theta = 2\sin^2\left(\frac{\theta}{2}\right) \] \[ \sin\theta = 2\sin\left(\frac{\theta}{2}\right)\cos\left(\frac{\theta}{2}\right) \] Substitute these identities back into the expression for $z$: \[ z = 2\sin^2\left(\frac{\theta}{2}\right) + i \cdot 2\sin\left(\frac{\theta}{2}\right)\cos\left(\frac{\theta}{2}\right) = 2\sin\left(\frac{\theta}{2}\right) \left[ \sin\left(\frac{\theta}{2}\right) + i\cos\left(\frac{\theta}{2}\right) \right] \] To convert this into standard polar form, the real part must be a cosine function and the imaginary part must be a sine function. Use co-function identities to switch them: \[ \sin\left(\frac{\theta}{2}\right) = \cos\left(\frac{\pi}{2} - \frac{\theta}{2}\right) \quad \text{and} \quad \cos\left(\frac{\theta}{2}\right) = \sin\left(\frac{\pi}{2} - \frac{\theta}{2}\right) \] The expression becomes: \[ z = 2\sin\left(\frac{\theta}{2}\right) \left[ \cos\left(\frac{\pi}{2} - \frac{\theta}{2}\right) + i\sin\left(\frac{\pi}{2} - \frac{\theta}{2}\right) \right] \] Since $\theta \in (0, \pi)$, the amplitude term $2\sin(\frac{\theta}{2})$ is strictly positive, meaning the expression is in standard polar form. Therefore, the principal argument is $\frac{\pi - \theta}{2}$. Final Answer: $\frac{\pi - \theta}{2}$.

Since both terms are expressed in Euler's form, we can find their product quickly by multiplying the moduli and adding the exponents following standard exponential rules: \[ z = (2 \times 3) \cdot e^{i(\pi/3 + \pi/6)} \] Calculate the sum of the angles in the exponent: \[ \frac{\pi}{3} + \frac{\pi}{6} = \frac{2\pi + \pi}{6} = \frac{3\pi}{6} = \frac{\pi}{2} \] The expression simplifies to: \[ z = 6e^{i\pi/2} \] Now convert this back to standard form using Euler's formula ($e^{i\theta} = \cos\theta + i\sin\theta$): \[ z = 6\left(\cos\frac{\pi}{2} + i\sin\frac{\pi}{2}\right) \] Substitute the trigonometric values $\cos(\frac{\pi}{2}) = 0$ and $\sin(\frac{\pi}{2}) = 1$: \[ z = 6(0 + i(1)) = 6i \] Final Answer: $6i$.

Expand both complex exponential terms using Euler's identity: \[ e^{i\theta} = \cos\theta + i\sin\theta \] \[ e^{-i\theta} = \cos(-\theta) + i\sin(-\theta) = \cos\theta - i\sin\theta \] Substitute these expansions into the numerator of the expression for $w$: \[ e^{i\theta} + e^{-i\theta} = (\cos\theta + i\sin\theta) + (\cos\theta - i\sin\theta) \] The imaginary $+i\sin\theta$ and $-i\sin\theta$ terms cancel out completely: \[ e^{i\theta} + e^{-i\theta} = 2\cos\theta \] Divide the expression by 2: \[ w = \frac{2\cos\theta}{2} = \cos\theta \] This identity forms the foundation for hyperbolic functions and advanced calculus proofs. Final Answer: $\cos\theta$.

To evaluate a variable base raised to a complex exponent, we must first express the base unit $i$ in Euler's form. The complex number $i$ has a modulus of $1$ and a principal argument of $\frac{\pi}{2}$, so it can be written as: \[ i = 1 \cdot e^{i\pi/2} \] Substitute this conversion back into the expression for $E$: \[ E = (e^{i\pi/2})^i \] Apply standard exponential multiplication rules to the exponents: \[ E = e^{i^2 \cdot \pi/2} \] Since $i^2 = -1$, substitute this value into the exponent: \[ E = e^{-\pi/2} \] Notice that $e^{-\pi/2} \approx 0.207$ is a purely real number. This surprising result is a common conceptual question on the JEE Advanced exam. Final Answer: $e^{-\pi/2}$.

First, convert the terms in the numerator and denominator into standard form using De Moivre's Theorem or Euler's form: For the numerator: \[ (\cos 2\theta + i\sin 2\theta)^5 = (e^{i2\theta})^5 = e^{i10\theta} \] For the denominator, adjust the sign using the negative angle identities $\cos(-\phi) = \cos\phi$ and $\sin(-\phi) = -\sin\phi$: \[ \cos 3\theta - i\sin 3\theta = \cos(-3\theta) + i\sin(-3\theta) = e^{-i3\theta} \] Now raise this expression to the third power as indicated in the denominator: \[ (e^{-i3\theta})^3 = e^{-i9\theta} \] Substitute these simplified terms back into the fraction expression $E$: \[ E = \frac{e^{i10\theta}}{e^{-i9\theta}} \] Combine the exponents following standard division rules: \[ E = e^{i10\theta} \cdot e^{i9\theta} = e^{i(10\theta + 9\theta)} = e^{i19\theta} \] Convert back to trigonometric form: \[ E = \cos 19\theta + i\sin 19\theta \] Final Answer: $\cos 19\theta + i\sin 19\theta$.

By De Moivre's Theorem, we know that: \[ \cos 3\theta + i\sin 3\theta = (\cos\theta + i\sin\theta)^3 \] Expand the right-hand side of the equation using the binomial theorem: \[ (\cos\theta + i\sin\theta)^3 = \binom{3}{0}\cos^3\theta + \binom{3}{1}\cos^2\theta(i\sin\theta) + \binom{3}{2}\cos\theta(i\sin\theta)^2 + \binom{3}{3}(i\sin\theta)^3 \] Evaluate the powers of the imaginary unit $i$ ($i^2 = -1$ and $i^3 = -i$): \[ = \cos^3\theta + 3i\cos^2\theta\sin\theta - 3\cos\theta\sin^2\theta - i\sin^3\theta \] Group the terms into real and imaginary parts: \[ = (\cos^3\theta - 3\cos\theta\sin^2\theta) + i(3\cos^2\theta\sin\theta - \sin^3\theta) \] To isolate $\cos 3\theta$, equate the real parts from both sides of the equation: \[ \cos 3\theta = \cos^3\theta - 3\cos\theta\sin^2\theta \] Use the Pythagorean identity $\sin^2\theta = 1 - \cos^2\theta$ to express the equation purely in terms of cosine: \[ \cos 3\theta = \cos^3\theta - 3\cos\theta(1 - \cos^2\theta) = \cos^3\theta - 3\cos\theta + 3\cos^3\theta \] Combine like terms to complete the derivation: \[ \cos 3\theta = 4\cos^3\theta - 3\cos\theta \] Final Answer: $4\cos^3\theta - 3\cos\theta$.

Convert the complex number base into its polar form first. Let $z = \frac{1}{\sqrt{2}} + i\frac{1}{\sqrt{2}}$: The modulus is $r = \sqrt{(\frac{1}{\sqrt{2}})^2 + (\frac{1}{\sqrt{2}})^2} = \sqrt{\frac{1}{2} + \frac{1}{2}} = 1$.
The argument angle is $\theta = \arctan(1/1) = \frac{\pi}{4}$. Thus, $z = e^{i\pi/4}$.
Now raise this expression to the $2026^{\text{th}}$ power: \[ z^{2026} = (e^{i\pi/4})^{2026} = e^{i \cdot \frac{2026\pi}{4}} = e^{i \cdot \frac{1013\pi}{2}} \] To simplify this large fraction, divide $1013$ by $2$: \[ \frac{1013\pi}{2} = 506\pi + \frac{\pi}{2} \] Since adding multiples of $2\pi$ corresponds to full rotations around the circle, they can be omitted without changing the value: \[ e^{i(506\pi + \pi/2)} = e^{i\pi/2} = i \] Final Answer: $i$.

Use the core cube root identity $1 + \omega + \omega^2 = 0$ to find expressions for the terms inside the parentheses: From the identity, we can isolate parts of the sum: \[ 1 + \omega^2 = -\omega \] \[ 1 + \omega = -\omega^2 \] Substitute these equations into the two factors of the expression $E$: \[ \text{First factor: } (1 + \omega^2) - \omega = (-\omega) - \omega = -2\omega \] \[ \text{Second factor: } (1 + \omega) - \omega^2 = (-\omega^2) - \omega^2 = -2\omega^2 \] Multiply the two simplified factors together: \[ E = (-2\omega) \times (-2\omega^2) = 4\omega^3 \] Now apply the second identity property $\omega^3 = 1$: \[ E = 4 \times 1 = 4 \] Final Answer: $4$.

The standard factorization of a sum of cubes is: \[ x^3 + y^3 = (x+y)(x^2 - xy + y^2) \] Let us focus on rewriting the quadratic factor $(x^2 - xy + y^2)$ using the properties of $\omega$. Recall that $1 = \omega^3$ and $-\omega - \omega^2 = 1 \implies -1 = \omega + \omega^2$. Substitute these identities into the coefficients of the quadratic expression: \[ x^2 + (-1)xy + (1)y^2 = x^2 + (\omega + \omega^2)xy + (\omega^3)y^2 \] Expand the terms: \[ = x^2 + \omega xy + \omega^2 xy + \omega^3 y^2 \] Group the terms to factor the expression by grouping: \[ = x(x + \omega y) + \omega^2 y(x + \omega y) = (x + \omega y)(x + \omega^2 y) \] Combine this result back with the first linear factor to find the complete factorization: \[ x^3 + y^3 = (x+y)(x + \omega y)(x + \omega^2 y) \] This technique can be used similarly to factor $x^3 - y^3$ and $x^3 + y^3 + z^3 - 3xyz$. Final Answer: $(x+y)(x + \omega y)(x + \omega^2 y)$.

Since $\omega^3 = 1$, the powers of $\omega$ repeat in a cycle of $3$. To simplify the expressions, divide the exponents by $3$ and find the remainders: For the first term: \[ 2026 = 3 \times 675 + 1 \implies \omega^{2026} = (\omega^3)^{675} \cdot \omega^1 = 1 \cdot \omega = \omega \] For the second term: \[ 4052 = 3 \times 1350 + 2 \implies \omega^{4052} = (\omega^3)^{1350} \cdot \omega^2 = 1 \cdot \omega^2 = \omega^2 \] Substitute these simplified terms back into the sum $S$: \[ S = \omega + \omega^2 \] Apply the core identity $1 + \omega + \omega^2 = 0 \implies \omega + \omega^2 = -1$: \[ S = -1 \] Final Answer: $-1$.

The equation $z^5 - 1 = 0 \implies z^5 = 1$ defines the $5^{\text{th}}$ roots of unity, so $n = 5$.
According to the product property for the $n^{\text{th}}$ roots of unity: \[ \text{Product} = (-1)^{n+1} \] Substitute $n = 5$ into the exponent formula: \[ \text{Product} = (-1)^{5+1} = (-1)^6 = 1 \] Let us verify this result using the theory of equations. For a polynomial equation $z^5 + 0z^4 + 0z^3 + 0z^2 + 0z - 1 = 0$, the product of the roots is given by $(-1)^5 \times \frac{\text{constant term}}{\text{leading coefficient}}$: \[ \text{Product} = (-1) \times \frac{-1}{1} = 1 \] Both methods yield the exact same result. Final Answer: $1$.

The roots of the equation $z^n - 1 = 0$ are $1, \alpha_1, \alpha_2, \dots, \alpha_{n-1}$. We can write the polynomial as a product of its linear root factors: \[ z^n - 1 = (z - 1)(z - \alpha_1)(z - \alpha_2)\dots(z - \alpha_{n-1}) \] Divide both sides of the equation by the linear factor $(z - 1)$: \[ \frac{z^n - 1}{z - 1} = (z - \alpha_1)(z - \alpha_2)\dots(z - \alpha_{n-1}) \] Apply the algebraic identity for the sum of a geometric series to rewrite the left-hand side: \[ z^{n-1} + z^{n-2} + \dots + z + 1 = (z - \alpha_1)(z - \alpha_2)\dots(z - \alpha_{n-1}) \] To evaluate the target product expression $P$, evaluate this polynomial identity at the point $z = 1$: \[ 1^{n-1} + 1^{n-2} + \dots + 1 + 1 = (1 - \alpha_1)(1 - \alpha_2)\dots(1 - \alpha_{n-1}) \] The left-hand side is a sum of the number $1$ repeated exactly $n$ times. Therefore: \[ n = (1 - \alpha_1)(1 - \alpha_2)\dots(1 - \alpha_{n-1}) \] This identity is a powerful shortcut for solving root product problems on the JEE Advanced exam. Final Answer: $n$.

• For $k = 0$: $z_0 = 2e^{i\pi/6}$
• For $k = 1$: $z_1 = 2e^{i(5\pi/6)}$
• For $k = 2$: $z_2 = 2e^{i(9\pi/6)} = 2e^{i3\pi/2} = 2e^{-i\pi/2}$

Let us rearrange the terms inside the modulus to match the standard distance layout format $|z - z_0| = r$: \[ |z - (3 - i)| = 4 \] This equation states that the geometric distance between the moving point $z$ and the fixed point $z_0 = 3 - i$ is always exactly equal to $4$.
This matches the definition of a circle. The center of the circle is located at the coordinates $(3, -1)$ on the Argand plane, and its radius is exactly $4$. We can verify this by substituting $z = x + iy$ into the equation: \[ |(x-3) + i(y+1)| = 4 \implies \sqrt{(x-3)^2 + (y+1)^2} = 4 \] Square both sides of the equation: \[ (x-3)^2 + (y+1)^2 = 16 \] This is the standard Cartesian equation of a circle. Final Answer: A circle centered at $3 - i$ with a radius of $4$.

The equation is of the form $|z - z_1| = |z - z_2|$, where $z_1 = 2i$ (coordinates $(0, 2)$) and $z_2 = 4$ (coordinates $(4, 0)$). Geometrically, this represents the perpendicular bisector of the line segment connecting these two points.
Let us find the equation of this line using Cartesian coordinates by substituting $z = x + iy$: \[ |x + i(y-2)| = |(x-4) + iy| \] Calculate the modulus expressions on both sides: \[ \sqrt{x^2 + (y-2)^2} = \sqrt{(x-4)^2 + y^2} \] Square both sides to eliminate the radicals: \[ x^2 + y^2 - 4y + 4 = x^2 - 8x + 16 + y^2 \] Cancel out the quadratic terms $x^2$ and $y^2$ from both sides: \[ -4y + 4 = -8x + 16 \] Rearrange the terms into standard linear form: \[ 8x - 4y - 12 = 0 \implies 2x - y - 3 = 0 \] This is the equation of a straight line. Final Answer: A straight line with the equation $2x - y - 3 = 0$.

The equation is of the form $|z - z_1| + |z - z_2| = 2a$, where $z_1 = 2$, $z_2 = -2$, and $2a = 6 \implies a = 3$.
This definition matches an ellipse, provided the total sum value ($2a = 6$) is strictly greater than the distance between the two fixed foci points ($|z_1 - z_2|$): \[ \text{Distance between foci} = |2 - (-2)| = |4| = 4 \] Since $6 > 4$, the condition is met, and the equation represents a valid ellipse. The foci are located at $(2, 0)$ and $(-2, 0)$, and the length of the major axis is $6$. Final Answer: An ellipse with foci at $2$ and $-2$, and a major axis of length $6$.

To find the logarithm using the formula $\ln z = \ln|z| + i\arg(z)$, calculate the modulus and principal argument of $z$: Step 1: Calculate the modulus $|z|$: \[ |z| = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = 2 \] Step 2: Calculate the principal argument $\arg(z)$. The point lies in the first quadrant: \[ \theta = \arctan\left(\frac{\sqrt{3}}{1}\right) = \frac{\pi}{3} \] Step 3: Substitute these values into the complex logarithm formula: \[ \ln z = \ln(2) + i\frac{\pi}{3} \] This separates the logarithm cleanly into its real part $\ln(2)$ and its imaginary part $\frac{\pi}{3}$. Final Answer: $\ln 2 + i\frac{\pi}{3}$.

The complex number is of the form $a + ib$, where $a = 7$ and $b = 24$. Since $b = 24 > 0$, the real and imaginary parts of the square root must have the same sign.
Step 1: Calculate the absolute modulus $|z|$: \[ |z| = \sqrt{7^2 + 24^2} = \sqrt{49 + 576} = \sqrt{625} = 25 \] Step 2: Apply the standard formulas to find the real part $x$ and imaginary part $y$ of the root: \[ x = \sqrt{\frac{|z| + a}{2}} = \sqrt{\frac{25 + 7}{2}} = \sqrt{\frac{32}{2}} = \sqrt{16} = 4 \] \[ y = \sqrt{\frac{|z| - a}{2}} = \sqrt{\frac{25 - 7}{2}} = \sqrt{\frac{18}{2}} = \sqrt{9} = 3 \] Step 3: Combine the components, using the same sign for both parts because $b > 0$: \[ \sqrt{7 + 24i} = \pm(4 + 3i) \] Let us verify this result by squaring the answer: $(4+3i)^2 = 16 + 24i - 9 = 7 + 24i$. The solution is confirmed. Final Answer: $\pm(4 + 3i)$.

Convert the real number base $-1$ into its polar form on the Argand plane: The point lies on the negative real axis, so its modulus is $|-1| = 1$ and its principal argument angle is $\arg(-1) = \pi$. Write the number in exponential form: \[ -1 = 1 \cdot e^{i\pi} \] Apply the complex logarithm formula to this expression: \[ \ln(-1) = \ln(1) + i\pi \] Since $\ln(1) = 0$, the real part vanishes, leaving only the imaginary component: \[ \ln(-1) = i\pi \] The imaginary part of this expression is exactly $\pi$. Final Answer: $\pi$.