Complex number operations

A complex number in algebraic (Cartesian) form is

where: $a=\operatorname{Re}(z)$ is the real part and $b=\operatorname{Im}(z)$ is the imaginary part.

Addition and subtraction combine real parts with real parts and imaginary parts with imaginary parts:

Multiplication expands like any binomial product, then replaces $i^2$ with $-1$:

Conjugate and modulus are the tools that make division possible. The conjugate of $z=a+ib$ is $\bar z=a-ib$, and multiplying a complex number by its own conjugate always clears the imaginary part: $z\bar z=a^2+b^2$, a non-negative real number. The modulus is

The multiplicative inverse for $z\neq0$ is $z^{-1}=\bar z/|z|^2$, and the additive inverse is simply $-z=-a-ib$.

Equality of complex numbers. Two complex numbers $a+ib$ and $c+id$ are equal only when $a=c$ and $b=d$ — the real parts match and the imaginary parts match independently, as two separate conditions rather than one. This is why "equate real and imaginary parts" is a legitimate algebraic move: it is not a shortcut, it is the definition of equality for complex numbers, and it is used whenever an unknown complex number is compared against a known target.

Powers of $i$. Because $i^2=-1$, every higher power of $i$ cycles through just four values:

and then the pattern repeats: $i^5=i$, $i^6=-1$, and so on. To simplify $i^n$ for a large exponent $n$, divide $n$ by $4$ and keep only the remainder, since $i^n=i^{\,n\bmod 4}$ (a remainder of $0$ gives $i^0=1$). Memorising this four-value cycle is faster than recomputing powers of $i$ from scratch every time one appears in an expansion. The svg below shows $z=a+ib$ as a point on the Argand plane, with modulus $r=|z|$ the distance from the origin and argument $\theta$ the angle from the positive real axis, giving the polar form $z=r(\cos\theta+i\sin\theta)$.

Converting to polar form means finding $r=|z|$ and $\theta$ from $a$ and $b$. Take $z=-1+i\sqrt3$: here $a=-1$, $b=\sqrt3$, so $r=\sqrt{(-1)^2+(\sqrt3)^2}=\sqrt{1+3}=2$. Then $\cos\theta=a/r=-1/2$ and $\sin\theta=b/r=\sqrt3/2$; since $\cos\theta<0$ and $\sin\theta>0$, $\theta$ lies in the second quadrant, so $\theta=120^\circ=2\pi/3$. Both signs must be checked together — reading off a reference angle from $\sin\theta$ alone, ignoring the sign of $\cos\theta$, is the single commonest slip in this step. Putting the pieces together, $z=2\left(\cos\tfrac{2\pi}{3}+i\sin\tfrac{2\pi}{3}\right)$, which can be checked by expanding back: $2\cos120^\circ=2(-1/2)=-1$ and $2\sin120^\circ=2(\sqrt3/2)=\sqrt3$, matching $a$ and $b$ exactly.

Let $z_1=3+4i$ and $z_2=1-2i$. Find $z_1+z_2$, $z_1z_2$, and $z_1/z_2$.

Step 1. Sum: real parts and imaginary parts combine separately, $z_1+z_2=(3+1)+i(4-2)=4+2i$.

Step 2. Product: expand as a binomial product and replace $i^2$ with $-1$,

Step 3. Quotient: multiply by the conjugate of the denominator, $1+2i$,

So $z_1+z_2=4+2i$, $z_1z_2=11-2i$, and $z_1/z_2=-1+2i$.

Competitive stretch

Find $|z_1|$, the polar form of $z_1=3+4i$, and $\sqrt{3+4i}$.

Step 1. Modulus: $|z_1|=\sqrt{3^2+4^2}=\sqrt{9+16}=\sqrt{25}=5$.

Step 2. Square root: let $\sqrt{3+4i}=x+iy$. Squaring both sides and equating real and imaginary parts gives $x^2-y^2=3$ and $2xy=4$, so $xy=2$. Also $x^2+y^2=|3+4i|=5$ (a shortcut that avoids solving a quartic directly).

Step 3. Add and subtract the two equations $x^2+y^2=5$ and $x^2-y^2=3$: adding gives $2x^2=8\Rightarrow x^2=4\Rightarrow x=\pm2$; subtracting gives $2y^2=2\Rightarrow y^2=1\Rightarrow y=\pm1$. Since $xy=2>0$, $x$ and $y$ must share a sign, so $(x,y)=(2,1)$ or $(-2,-1)$, giving $\sqrt{3+4i}=\pm(2+i)$.

Step 4. Check by squaring: $(2+i)^2=4+4i+i^2=4+4i-1=3+4i$ — confirmed. The modulus is $|z_1|=5$, and $\sqrt{3+4i}=\pm(2+i)$. The square root of a complex number always comes as a $\pm$ pair, exactly like real square roots, because if $w^2=z$ then $(-w)^2=z$ too.

• Dividing by multiplying by the conjugate of the numerator instead of the denominator — it must always be the denominator's conjugate.
• Forgetting $i^2=-1$ carries through every expansion, not just the first occurrence of $i^2$.
• Writing $\sqrt{-a}\times\sqrt{-b}=\sqrt{ab}$ for positive $a,b$ — convert to $i$-form first ($\sqrt{-a}=i\sqrt a$) before multiplying negative radicands.
• Confusing modulus (a non-negative real number $|z|$) with argument (an angle $\theta$) — they answer different questions.
• Reading off the argument's quadrant from $\cos\theta$ or $\sin\theta$ alone — always check the signs of both the real and imaginary parts together before deciding which quadrant $\theta$ lies in.

These five traps account for most of the marks lost on this topic in practice — each one is a small, avoidable slip rather than a gap in understanding, so a quick self-check against this list before submitting an answer is worth the ten seconds it takes.

Ask the tutor: if any step of rationalising a quotient or expanding a product with $i$ feels shaky, ask the AI tutor to redo it with fresh numbers or check your working line by line.