Complex Numbers and Quadratic Equations • Topic 1 of 3

Complex Numbers and Their Algebra

Some equations have no real solution. The simplest is $x^2 = -1$, because the square of any real number is never negative. To repair this gap we introduce a single new symbol, the imaginary unit $i$, defined by the one rule that does all the work:

$$i = \sqrt{-1}, \qquad i^2 = -1$$

A complex number is any expression of the form $z = a + ib$, where $a$ and $b$ are real numbers. Here $a$ is the real part, written $\operatorname{Re}(z)$, and $b$ is the imaginary part, written $\operatorname{Im}(z)$. Notice that $\operatorname{Im}(z) = b$ is the coefficient of $i$, not $ib$ itself. When $b = 0$ the number is purely real, and when $a = 0$ (with $b \ne 0$) it is purely imaginary.

Equality: two complex numbers are equal exactly when their real parts match and their imaginary parts match — a single equation in complex numbers is really two equations in disguise.

$$a + ib = c + id \iff a = c \text{ and } b = d$$

Powers of $i$ cycle every four steps. From $i^2 = -1$ we get $i^3 = i^2 \cdot i = -i$ and $i^4 = (i^2)^2 = 1$. After that the pattern repeats, so to find any power of $i$ you only need the remainder of the exponent on division by $4$.

PowerValuePowerValue
$i^1$$i$$i^5$$i$
$i^2$$-1$$i^6$$-1$
$i^3$$-i$$i^7$$-i$
$i^4$$1$$i^8$$1$

Arithmetic. Addition and subtraction act part by part. Multiplication uses ordinary distribution, after which every $i^2$ is replaced by $-1$ and like terms are gathered:

$$(a + ib) + (c + id) = (a + c) + i(b + d)$$
$$(a + ib)(c + id) = (ac - bd) + i(ad + bc)$$

Division is the one operation that needs a trick: multiply the top and bottom by the conjugate of the denominator, $c - id$, which turns the denominator into the real number $c^2 + d^2$.

$$\dfrac{a + ib}{c + id} = \dfrac{(a + ib)(c - id)}{c^2 + d^2}$$

A caution with square roots of negatives: the schoolbook rule $\sqrt{x}\,\sqrt{y} = \sqrt{xy}$ fails when both numbers are negative. Always convert first: $\sqrt{-4} = 2i$, so $\sqrt{-4}\,\sqrt{-9} = (2i)(3i) = 6i^2 = -6$, never $\sqrt{36} = 6$.

Deeper Insight — one symbol, and the whole number system closes up: The entire chapter rests on a single act of definition: declare $i^2 = -1$ and agree that $i$ obeys the ordinary laws of algebra. Everything else is forced. Once you accept $a + ib$, addition and multiplication are not new rules to memorise but the familiar distributive law applied while remembering to swap $i^2$ for $-1$. The deeper pay-off is closure: in the real numbers, polynomial equations could run off the edge of the system and have no answer, but the moment $i$ exists, every polynomial equation has a full set of roots — this is the Fundamental Theorem of Algebra, and it is why complex numbers feel less like an invention and more like a discovery. Treat $i$ as an ordinary quantity you can add and multiply, never as something mystical, and the algebra behaves exactly as you expect.

Anatomy of a complex number a plus ib Anatomy of a Complex Number z = a + i b real part imaginary part a and b are real numbers, i = √(−1) The four-step cycle of powers of i Powers of i Repeat Every 4 Steps i−1−i1 multiply by ito step round
1
Worked Example
Express $(3 + 2i) + (5 - 7i)$ in the form $a + ib$.
Solution
  1. Add the real parts: $3 + 5 = 8$.
  2. Add the imaginary parts: $2 + (-7) = -5$.
  3. Combine into standard form.

Answer: $8 - 5i$

2
Worked Example
Simplify $i^{73}$.
Solution
  1. Powers of $i$ repeat every $4$, so divide the exponent by $4$: $73 = 4 \times 18 + 1$.
  2. The remainder is $1$, so $i^{73} = i^1$.

Answer: $i^{73} = i$

3
Worked Example
Multiply $(2 + 3i)(4 - 5i)$ and write the result as $a + ib$.
Solution
  1. Expand by distribution: $2(4) + 2(-5i) + 3i(4) + 3i(-5i)$.
  2. $= 8 - 10i + 12i - 15i^2$.
  3. Replace $i^2$ with $-1$: $-15i^2 = +15$.
  4. Collect: $(8 + 15) + (-10 + 12)i$.

Answer: $23 + 2i$

4
Worked Example
Express $\dfrac{3 + 2i}{1 - 4i}$ in the form $a + ib$.
Solution
  1. Multiply numerator and denominator by the conjugate of the denominator, $1 + 4i$.
  2. Denominator: $(1 - 4i)(1 + 4i) = 1 - 16i^2 = 1 + 16 = 17$.
  3. Numerator: $(3 + 2i)(1 + 4i) = 3 + 12i + 2i + 8i^2 = 3 + 14i - 8 = -5 + 14i$.
  4. Divide each part by $17$.

Answer: $-\dfrac{5}{17} + \dfrac{14}{17}i$

5
Worked Example
Find real numbers $x$ and $y$ such that $(x + 2y) + i(2x - y) = 5 - i$.
Solution
  1. Equate real parts: $x + 2y = 5$.
  2. Equate imaginary parts: $2x - y = -1$.
  3. From the second, $y = 2x + 1$. Substitute: $x + 2(2x + 1) = 5 \Rightarrow 5x + 2 = 5 \Rightarrow x = \dfrac{3}{5}$.
  4. Then $y = 2\left(\dfrac{3}{5}\right) + 1 = \dfrac{11}{5}$.

Answer: $x = \dfrac{3}{5},\ y = \dfrac{11}{5}$

6
Worked Example
Evaluate $\sqrt{-16}\,\cdot\,\sqrt{-25}$.
Solution
  1. Convert each root before multiplying — the rule $\sqrt{x}\sqrt{y} = \sqrt{xy}$ fails for two negatives.
  2. $\sqrt{-16} = 4i$ and $\sqrt{-25} = 5i$.
  3. Multiply: $(4i)(5i) = 20i^2 = 20(-1) = -20$.

Answer: $-20$

Key Points

  • The imaginary unit satisfies $i^2 = -1$; a complex number has the standard form $z = a + ib$ with $a, b \in \mathbb{R}$.
  • Powers of $i$ cycle with period $4$ — reduce the exponent modulo $4$ to evaluate $i^n$.
  • Add and subtract part by part; multiply by distribution and replace every $i^2$ with $-1$.
  • Divide by multiplying numerator and denominator by the conjugate of the denominator.
  • Always rewrite $\sqrt{-k}$ as $\sqrt{k}\,i$ before combining roots of negative numbers.
Tap an option to check your answer0 / 4
Q1.$i^2=$
Explanation: By definition $i^2=-1$.
Q2.$(2+3i)+(1-i)=$
Explanation: Add real and imaginary parts separately.
Q3.$i^4=$
Explanation: $i^4=(i^2)^2=1$.
Q4.$(1+i)(1-i)=$
Explanation: $=1-i^2=1+1=2$.