Mathematics 9709 · Pure Mathematics 3

Polar and exponential form; multiply and divide

Once you know the length $r=|z|$ and the angle $\theta=\arg z$ , you can write $z$ in two more ways.
Both use the same two numbers $(r,\theta)$ . And both make multiplying complex numbers very easy.

In polar form, multiplying is easy: multiply the lengths, add the angles. To divide: divide the lengths, subtract the angles. No expanding brackets, no conjugates.

Two new ways to write the same z

From $x=r\cos\theta$ and $y=r\sin\theta$ , put these back into $z=x+iy$ :

$z=r(\cos\theta+i\sin\theta)=re^{i\theta}$

The middle one is modulus–argument form; the right one is exponential form (Euler's result $\cos\theta+i\sin\theta=e^{i\theta}$ ).
In both, $r=|z|$ and $\theta=\arg z$ : the same values you found in §11.3, so there is nothing new to work out.

z = 3 + 3i has length r = 3√2 and angle θ = π/4, so z = 3√2(cos π/4 + i sin π/4) = 3√2 e^(iπ/4).

Multiply: ×lengths, +angles. Divide: ÷lengths, −angles

Because $re^{i\theta}$ follows the index laws, for $z_1=r_1e^{i\theta_1}$ and $z_2=r_2e^{i\theta_2}$ :

$z_1z_2=r_1r_2\,e^{i(\theta_1+\theta_2)},\qquad \dfrac{z_1}{z_2}=\dfrac{r_1}{r_2}\,e^{i(\theta_1-\theta_2)}$

So $|z_1z_2|=|z_1||z_2|$ and $\arg(z_1z_2)=\arg z_1+\arg z_2$ .

Why multiplying adds the angles

Multiply two exponentials and the index law just adds the powers: $r_1e^{i\theta_1}\cdot r_2e^{i\theta_2}=r_1r_2\,e^{i(\theta_1+\theta_2)}$ . On the diagram, multiplying by $z_2$ turns $z_1$ by $\arg z_2$ and changes its length by a factor of $|z_2|$ . So multiplying by $i$ (which is $e^{i\pi/2}$ ) turns a point a quarter-turn anticlockwise. This is the easiest way to see $i^2=-1$ : two quarter-turns make a half-turn, which sends $1$ to $-1$ .

Worked examples

Worked example

Write $1+i$ in both polar forms

Modulus: $r=\sqrt{1^2+1^2}=\sqrt2$ .
First quadrant, so the argument is the reference angle itself: $\theta=\tan^{-1}1=\tfrac{\pi}{4}$ .

Answer

\sqrt2\big(\cos\tfrac{\pi}{4}+i\sin\tfrac{\pi}{4}\big)=\sqrt2\,e^{i\pi/4}

Worked example

Real question: $u=-4+2i$ , $v=3+i$ , write $u/v$ in exact form $re^{i\theta}$

9709/32 Feb/March 2021 Q8(b), right after the division in §11.2.

From part (a), $\dfrac{u}{v}=-1+i$ .
Modulus: $r=\sqrt{(-1)^2+1^2}=\sqrt2$ .
The point $-1+i$ is in the second quadrant, so $\theta=\pi-\tfrac{\pi}{4}=\tfrac{3\pi}{4}$ .

Answer

\dfrac{u}{v}=\sqrt2\,e^{i3\pi/4}

Worked example

Real question: $z=-3+i$ , write $z^2$ in the form $re^{i\theta}$

9709/32 Feb/March 2024 Q3(a). Square it in x + iy form first, then convert.

Square it: $z^2=(-3+i)^2=9-6i+i^2=8-6i$ .
Modulus: $r=\sqrt{8^2+(-6)^2}=\sqrt{100}=10$ .
$8-6i$ is in the fourth quadrant (right, down), so $\theta=-\tan^{-1}\tfrac{6}{8}=-0.6435$ .

Answer

z^2=10\,e^{-0.644\,i}

Where the marks go

Common mistake

After adding or subtracting arguments, pull the answer back into the principal range

(-\pi,\pi]

: if the sum goes above

\pi

, take away

2\pi

; if it reaches or drops below

-\pi

, add

2\pi

. And keep

\theta

in radians unless told otherwise.

Common mistake

“Exact” means leave it as

\sqrt2\,e^{i3\pi/4}

, not a rounded decimal. Use polar form when a question says multiply, divide or raise to a power; for a single add or subtract, x + iy form is quicker. (9709/32 Feb/March 2021 Q8)

Now you try

Given $u=2e^{i\pi/3}$ and $w=3e^{i\pi/2}$ , find $uw$ and $\dfrac{u}{w}$ in exponential form. (9709-style)

Your turn— tap to reveal the worked answer (9709-style)

For $uw$ : multiply the lengths, add the angles: $uw=6e^{i(5\pi/6)}$ . For $\tfrac{u}{w}$ : divide the lengths, subtract the angles: $\tfrac{u}{w}=\tfrac23 e^{-i\pi/6}$ .

Answer

uw=6e^{i5\pi/6},\quad \tfrac{u}{w}=\tfrac23 e^{-i\pi/6}

PreviousThe Argand diagram: modulus and argument NextSolving equations with complex roots

Polar and exponential form; multiply and divide

Once you know the length $r=|z|$ and the angle $\theta=\arg z$ , you can write $z$ in two more ways.
Both use the same two numbers $(r,\theta)$ . And both make multiplying complex numbers very easy.

In polar form, multiplying is easy: multiply the lengths, add the angles. To divide: divide the lengths, subtract the angles. No expanding brackets, no conjugates.

Two new ways to write the same z

From $x=r\cos\theta$ and $y=r\sin\theta$ , put these back into $z=x+iy$ :

$z=r(\cos\theta+i\sin\theta)=re^{i\theta}$

The middle one is modulus–argument form; the right one is exponential form (Euler's result $\cos\theta+i\sin\theta=e^{i\theta}$ ).
In both, $r=|z|$ and $\theta=\arg z$ : the same values you found in §11.3, so there is nothing new to work out.

z = 3 + 3i has length r = 3√2 and angle θ = π/4, so z = 3√2(cos π/4 + i sin π/4) = 3√2 e^(iπ/4).

Multiply: ×lengths, +angles. Divide: ÷lengths, −angles

Because $re^{i\theta}$ follows the index laws, for $z_1=r_1e^{i\theta_1}$ and $z_2=r_2e^{i\theta_2}$ :

$z_1z_2=r_1r_2\,e^{i(\theta_1+\theta_2)},\qquad \dfrac{z_1}{z_2}=\dfrac{r_1}{r_2}\,e^{i(\theta_1-\theta_2)}$

So $|z_1z_2|=|z_1||z_2|$ and $\arg(z_1z_2)=\arg z_1+\arg z_2$ .

Why multiplying adds the angles

Worked examples

Worked example

Write $1+i$ in both polar forms

Modulus: $r=\sqrt{1^2+1^2}=\sqrt2$ .
First quadrant, so the argument is the reference angle itself: $\theta=\tan^{-1}1=\tfrac{\pi}{4}$ .

Answer

\sqrt2\big(\cos\tfrac{\pi}{4}+i\sin\tfrac{\pi}{4}\big)=\sqrt2\,e^{i\pi/4}

Worked example

Real question: $u=-4+2i$ , $v=3+i$ , write $u/v$ in exact form $re^{i\theta}$

9709/32 Feb/March 2021 Q8(b), right after the division in §11.2.

From part (a), $\dfrac{u}{v}=-1+i$ .
Modulus: $r=\sqrt{(-1)^2+1^2}=\sqrt2$ .
The point $-1+i$ is in the second quadrant, so $\theta=\pi-\tfrac{\pi}{4}=\tfrac{3\pi}{4}$ .

Answer

\dfrac{u}{v}=\sqrt2\,e^{i3\pi/4}

Worked example

Real question: $z=-3+i$ , write $z^2$ in the form $re^{i\theta}$

9709/32 Feb/March 2024 Q3(a). Square it in x + iy form first, then convert.

Square it: $z^2=(-3+i)^2=9-6i+i^2=8-6i$ .
Modulus: $r=\sqrt{8^2+(-6)^2}=\sqrt{100}=10$ .
$8-6i$ is in the fourth quadrant (right, down), so $\theta=-\tan^{-1}\tfrac{6}{8}=-0.6435$ .

Answer

z^2=10\,e^{-0.644\,i}

Where the marks go

Common mistake

After adding or subtracting arguments, pull the answer back into the principal range

(-\pi,\pi]

: if the sum goes above

\pi

, take away

2\pi

; if it reaches or drops below

-\pi

, add

2\pi

. And keep

\theta

in radians unless told otherwise.

Common mistake

“Exact” means leave it as

\sqrt2\,e^{i3\pi/4}

, not a rounded decimal. Use polar form when a question says multiply, divide or raise to a power; for a single add or subtract, x + iy form is quicker. (9709/32 Feb/March 2021 Q8)

Now you try

Given $u=2e^{i\pi/3}$ and $w=3e^{i\pi/2}$ , find $uw$ and $\dfrac{u}{w}$ in exponential form. (9709-style)

Your turn— tap to reveal the worked answer (9709-style)

For $uw$ : multiply the lengths, add the angles: $uw=6e^{i(5\pi/6)}$ . For $\tfrac{u}{w}$ : divide the lengths, subtract the angles: $\tfrac{u}{w}=\tfrac23 e^{-i\pi/6}$ .

Answer

uw=6e^{i5\pi/6},\quad \tfrac{u}{w}=\tfrac23 e^{-i\pi/6}