Complex Numbers and Polar Form: From Giving Directions to Modulus and Argument

Complex numbers and polar form are two address systems for the same point. The rectangular form $z=x+yi$ says "across $x$, up $y$". In the language of complex numbers and polar form, the distance from the origin is the modulus $r$, and the angle from the positive real axis is the argument $\theta$.

Ground one number to see how complex numbers and polar form connect: the point $\sqrt{3}+i$, across $\sqrt{3}\approx1.73$, up 1.

Distance first, Pythagoras: $r=\sqrt{(\sqrt{3})^2+1^2}=\sqrt{4}=2$.

Direction next — the second coordinate of complex numbers and polar form. The legs are $\sqrt{3}$ and 1 with hypotenuse 2, so $\sin\theta=\frac{1}{2}$ and $\theta=30^{\circ}$.

In letters, the conversion kit for complex numbers and polar form is two pairs of formulas (same pairs as Byerly and Peirce, 1888):

• $r=\sqrt{x^2+y^2}$ and $\tan\theta=\frac{y}{x}$;
• $x=r\cos\theta$ and $y=r\sin\theta$.

Substitute the second pair into $z=x+yi$:

$z=x+yi=r\cos\theta+(r\sin\theta)i=r(\cos\theta+i\sin\theta)$

That last line is the heart of complex numbers and polar form: modulus $r$, argument $\theta$. Check on our point, because converting complex numbers and polar form must return the start: $2(\cos30^{\circ}+i\sin30^{\circ})=\sqrt{3}+i$. Two addresses, one point — complex numbers and polar form in a single picture.

Rectangular to polar — two steps. Converting complex numbers and polar form in this direction starts with the modulus $r=\sqrt{x^2+y^2}$. Then fix the argument: plot the point, note the quadrant, and use $\tan\theta=y/x$ for the reference angle.

Step two of complex numbers and polar form is where grades die. The $\tan^{-1}$ button only returns fourth-quadrant-style angles. In QII, take $180^{\circ}$ minus the reference angle; in QIII add $180^{\circ}$; in QIV keep the negative angle. The 1888 rule still holds: choose the angle that brings the point into the right quadrant — the one non-negotiable of complex numbers and polar form.

Polar to rectangular — two steps. Evaluate $\cos\theta$ and $\sin\theta$, then multiply through by $r$: $z=r\cos\theta+(r\sin\theta)i$. This direction of complex numbers and polar form is pure distribution.

Four special arguments live on the axes (an 1888 exercise): positive reals have argument $0$, negative reals $\pi$, positive pure imaginaries $\frac{\pi}{2}$, negative pure imaginaries $-\frac{\pi}{2}$. Axis points are the fast lane of complex numbers and polar form — learn them cold.

Problem. First direction of complex numbers and polar form: write $z=-4+4i$ in polar form.

Solution. Step 1, the modulus:

$r=\sqrt{(-4)^2+4^2}=\sqrt{32}=4\sqrt{2}\approx5.66$

Step 2, the argument — the direction half of complex numbers and polar form. The point $(-4,4)$ sits in QII, so sketch it first.

Reference angle: $\tan^{-1}\left|\frac{4}{-4}\right|=\tan^{-1}(1)=45^{\circ}$.

QII adjustment: $\theta=180^{\circ}-45^{\circ}=135^{\circ}$.

Assemble the polar form:

$z=4\sqrt{2}(\cos135^{\circ}+i\sin135^{\circ})$

Answer: $r=4\sqrt{2}\approx5.66$, $\theta=135^{\circ}$.

Check by converting back — cheap insurance for complex numbers and polar form: $4\sqrt{2}\cos135^{\circ}=-4$ and $4\sqrt{2}\sin135^{\circ}=4$, which is $-4+4i$ again.

Feel the trap now: raw $\tan^{-1}(\frac{4}{-4})$ returns $-45^{\circ}$, a QIV angle pointing the wrong way. Every conversion between complex numbers and polar form ends with a quadrant check — that habit is the whole exam game of complex numbers and polar form.

Problem. Now the mirror direction of complex numbers and polar form: convert $z=12\left(\cos\frac{\pi}{6}+i\sin\frac{\pi}{6}\right)$ to rectangular form.

Solution. Recall the two values that bridge complex numbers and polar form: $\cos\frac{\pi}{6}=\frac{\sqrt{3}}{2}$ and $\sin\frac{\pi}{6}=\frac{1}{2}$.

Multiply through by 12 — distribution, the workhorse of complex numbers and polar form:

$z=12\cdot\frac{\sqrt{3}}{2}+\left(12\cdot\frac{1}{2}\right)i=6\sqrt{3}+6i$

Answer: $z=6\sqrt{3}+6i\approx10.39+6i$ — the round trip of complex numbers and polar form, complete.

Polar form reports "how far, which way"; rectangular form reports "across, up"; the trig functions translate. Two steps, one distribution — that is all this direction of complex numbers and polar form takes.

Take 1, multiply by $i$, then by $i$ again, and then by $i$ once more. The 1 has turned into $i$, the $i$ into $-1$, the $-1$ into $-i$, and the fourth multiplication comes home to 1. Each multiplication rotates the point 90° counterclockwise about the origin — the rotation that complex numbers and polar form are built on. Byerly and Peirce (1888) wrote it as a rule: the operation $\sqrt{-1}$ rotates the point about the origin through 90°, and one repetition rotates it another 90°.

Multiplying by $i$ keeps the length (the modulus stays 1) and turns the direction by 90°. Multiplying by any complex number works the same way: stretch the length by a factor, turn the direction through an angle. The modulus and the argument are not just a second address — they are the two quantities that multiplication actually operates on. That is why complex numbers and polar form make the rule short: multiply the moduli, add the arguments. The figure draws the four rotations as one closed loop.

Problem. Time for the operation that complex numbers and polar form make trivial. Let $z_1=2(\cos30^{\circ}+i\sin30^{\circ})$ and $z_2=3(\cos60^{\circ}+i\sin60^{\circ})$. Find $z_1z_2$ and convert it back to rectangular form.

In polar form the rule is one sentence — the payoff of complex numbers and polar form: multiply the moduli, add the arguments. Chrystal's Algebra (1904) states it as a theorem. The modulus of a product is the product of the moduli. Its amplitude (the older name for argument) is the sum of the amplitudes.

Apply it:

$z_1z_2=2\cdot3\left[\cos(30^{\circ}+60^{\circ})+i\sin(30^{\circ}+60^{\circ})\right]=6(\cos90^{\circ}+i\sin90^{\circ})$

The arguments add to $90^{\circ}$, so the product lands on the imaginary axis: $z_1z_2=6i$.

Answer: $z_1z_2=6(\cos90^{\circ}+i\sin90^{\circ})=6i$.

Verify by brute force, using $z_1=\sqrt{3}+i$ and $z_2=\frac{3}{2}+\frac{3\sqrt{3}}{2}i$:

$(\sqrt{3}+i)\left(\tfrac{3}{2}+\tfrac{3\sqrt{3}}{2}i\right)=\tfrac{3\sqrt{3}}{2}+\tfrac{9}{2}i+\tfrac{3}{2}i-\tfrac{3\sqrt{3}}{2}=6i$

The real parts cancel; 6i survives. Five lines of algebra, one line in polar form — complex numbers and polar form win this race, closing the loop from the block above: multiplication rotates.

Division is symmetric — the same machine of complex numbers and polar form run in reverse: divide the moduli, subtract the arguments ($z_2\neq0$).

1. Wrong-quadrant argument. For $-4+4i$ the calculator returns $\tan^{-1}(\frac{4}{-4})=-45^{\circ}$; copying it blindly is wrong. The point sits in QII and the argument is $135^{\circ}$. Sketch first, then compute — the golden rule of complex numbers and polar form.

2. Dropping the i. The form is $r(\cos\theta+i\sin\theta)$, not $r(\cos\theta+\sin\theta)$. Lose the $i$ and every product rule for complex numbers and polar form collapses.

3. Negative modulus. The modulus is a distance, so $r=\sqrt{x^2+y^2}\ge0$ always. Direction belongs to the argument, never to $r$ — the division of labor inside complex numbers and polar form.

4. Axis points from memory. A negative real has argument $180^{\circ}$, not $0^{\circ}$; $-i$ has argument $-90^{\circ}$ (or $270^{\circ}$), not $90^{\circ}$. Learn the four axis cases — the street signs of complex numbers and polar form.

5. Multiplying the arguments. Products of polar complex numbers multiply the moduli and add the arguments; quotients subtract them. Multiplying the angles instead is the fastest way to lose the point — complex numbers and polar form only pay off when the rule is right.