Recipe

Convert between rectangular and exponential form

Use when: you have z=a+bi and need z=re^{i\theta}, or you have z=re^{i\theta} and need a+bi

Goal: the same complex number rewritten in whichever of the two forms you didn't start with, with r>0 and \theta picked correctly by quadrant

Taught in Polar and exponential form, revisited

1. 1

   Look at what you're actually given: a rectangular number a+bi, or an exponential number re^{i\theta}. The two directions use different halves of this recipe, so this is the fork that decides where to start.

   • given a+bi — continue to Step 2 (find r and \theta).
   • given re^{i\theta} — skip ahead to Step 4 (expand with Euler's formula).

   Why: the modulus-argument pair and the exponential form are two spellings of the same data, so converting one way is just running the other direction's steps in reverse.

2. 2

   Compute r=|z|=\sqrt{a^2+b^2} directly from the real and imaginary parts.

   Example

   For z=3-3\sqrt3\,i: a=3, b=-3\sqrt3, so r=\sqrt{3^2+(3\sqrt3)^2}=\sqrt{9+27}=\sqrt{36}=6.

   Why: r is just z's distance from the origin, and the Pythagorean theorem gives that distance straight from the horizontal and vertical legs a and b.

3. 3

   Compute the reference angle from \tan(\text{ref})=|b/a|, then adjust its sign, and if needed subtract it from \pi, so the result lands in (-\pi,\pi] and matches the quadrant z actually sits in.

   Example

   Continuing with z=3-3\sqrt3\,i: a=3>0, so this is the first case. \tan(\text{ref})=\dfrac{3\sqrt3}{3}=\sqrt3, so the reference angle is \pi/3. Since b=-3\sqrt3<0, \theta=-\pi/3.

   • a>0 (quadrant I or IV) — \theta is the reference angle itself, positive if b>0 and negative if b<0.
   • a<0 (quadrant II or III) — \theta is \pi minus the reference angle if b>0, or the reference angle minus \pi if b<0.
   • a=0 — skip the reference-angle triangle entirely; \theta=\pi/2 if b>0 and \theta=-\pi/2 if b<0.

   ⚡ Shortcut

   When a=0 or b=0, z sits right on an axis and \theta is one of 0,\pi/2,\pi,-\pi/2, read off directly with no reference-angle triangle needed at all.

   Why: the reference angle alone only ever gives an acute angle; the quadrant is what tells you which of the four sign/offset combinations actually matches where z sits.

4. 4

   Going the other direction, start from z=re^{i\theta} and rewrite the exponential as \cos\theta+i\sin\theta using Euler's formula, still carrying the factor of r out front.

   Example

   For z=4e^{-i\pi/3}: z=4\Big(\cos(-\pi/3)+i\sin(-\pi/3)\Big).

   Why: Euler's formula is exactly the identity that lets \cos\theta+i\sin\theta and e^{i\theta} stand in for each other; nothing else in this step is new.

5. 5

   Look up (or compute) \cos\theta and \sin\theta, then multiply both by r to land in a+bi form.

   Example

   \cos(-\pi/3)=\tfrac12 and \sin(-\pi/3)=-\tfrac{\sqrt3}{2}, so z=4\left(\frac12-\frac{\sqrt3}{2}i\right)=2-2\sqrt3\,i.

   Why: distributing r across both the cosine term and the sine term is what turns the compact exponential form back into an explicit real part and imaginary part.