If z = √(3) + i 2, then find the value of z¹⁰¹ + z¹⁰³ + z¹⁰⁵ — JEE Mathematics

Express $z$ in polar / Euler form:

$$z = \cos\left(\frac{\pi}{6}\right) + i\sin\left(\frac{\pi}{6}\right) = e^{i\pi/6}$$

Notice that $z^3 = e^{i\pi/2} = i$. Let's factor out $z^{101}$ from the expression:

$$S = z^{101}(1 + z^2 + z^4)$$

Since $z = e^{i\pi/6}$, we have:

$$z^2 = e^{i\pi/3} = \frac{1}{2} + \frac{\sqrt{3}}{2}i$$

$$z^4 = e^{i2\pi/3} = -\frac{1}{2} + \frac{\sqrt{3}}{2}i$$

Substitute these values into the bracketed expression:

$$1 + z^2 + z^4 = 1 + \left(\frac{1}{2} + \frac{\sqrt{3}}{2}i\right) + \left(-\frac{1}{2} + \frac{\sqrt{3}}{2}i\right) = 1 + \sqrt{3}i$$

Now evaluate $z^{101}$:

$$z^{101} = z^{99} \cdot z^2 = (z^3)^{33} \cdot z^2 = i^{33} \cdot z^2 = i \cdot z^2$$

$$i \cdot z^2 = i \left(\frac{1}{2} + \frac{\sqrt{3}}{2}i\right) = -\frac{\sqrt{3}}{2} + \frac{1}{2}i$$

Now multiply $z^{101}$ and $(1 + z^2 + z^4)$:

$$S = \left(-\frac{\sqrt{3}}{2} + \frac{1}{2}i\right)(1 + \sqrt{3}i)$$

$$S = -\frac{\sqrt{3}}{2} - \frac{3}{2}i + \frac{1}{2}i - \frac{\sqrt{3}}{2} = -\sqrt{3} - i$$

Answer: $-\sqrt{3} - i$