Multiplying and dividing in polar form

In the product of two numbers in polar form the moduli multiply, and the angles add:

The proof below opens the brackets and uses the sum formulas of angles:

• $\displaystyle {(\cos\theta_1+i\sin\theta_1)(\cos\theta_2+i\sin\theta_2)}={(\cos\theta_1\cos\theta_2-\sin\theta_1\sin\theta_2)+i(\sin\theta_1\cos\theta_2+\cos\theta_1\sin\theta_2)}$.
• The real part is $\displaystyle \cos(\theta_1+\theta_2)$, and the imaginary part is $\displaystyle \sin(\theta_1+\theta_2)$.
• It remains to multiply by $\displaystyle r_1r_2$.

Example: $\displaystyle {3\operatorname{cis}\frac{\pi}{4}\cdot2\operatorname{cis}\frac{\pi}{12}}={6\operatorname{cis}\frac{\pi}{3}}={3+3\sqrt{3}i}$.

Dividing is the inverse of multiplying, so in a quotient the moduli divide and the angles subtract:

• The reciprocal: $\displaystyle {\frac{1}{r\operatorname{cis}\theta}}={\frac{1}{r}\operatorname{cis}(-\theta)}$.
• The conjugate: $\displaystyle {\overline{r\operatorname{cis}\theta}}={r\operatorname{cis}(-\theta)}$ – a reflection in the real axis.
• Example: $\displaystyle {\frac{8\operatorname{cis}\frac{5\pi}{6}}{2\operatorname{cis}\frac{\pi}{3}}}={4\operatorname{cis}\frac{\pi}{2}}={4i}$.

Multiplying $\displaystyle z$ by $\displaystyle {w}={r\operatorname{cis}\theta}$ adds $\displaystyle \theta$ to the angle of $\displaystyle z$ and multiplies its distance from the origin by $\displaystyle r$. So in the Gauss plane:

• multiplying by $\displaystyle \operatorname{cis}\theta$ rotates the point about the origin by the angle $\displaystyle \theta$, anticlockwise;
• multiplying by a real $\displaystyle r>0$ stretches by $\displaystyle r$ (shrinks, when $\displaystyle r<1$);
• multiplying by $\displaystyle {i}={\operatorname{cis}\frac{\pi}{2}}$ is a quarter turn: $\displaystyle {(4+i)\cdot i}={-1+4i}$.

The coordinate system below shows $\displaystyle {z}={4+i}$, $\displaystyle {iz}={-1+4i}$ and $\displaystyle {i^2z}={-z}={-4-i}$: each multiplication by $\displaystyle i$ turns a quarter turn.

To rotate $\displaystyle z$ about a point $\displaystyle z_0$ by the angle $\displaystyle \theta$: rotate the vector from $\displaystyle z_0$ to $\displaystyle z$, and add it to $\displaystyle z_0$:

The example below finds the vertices $\displaystyle C$ and $\displaystyle D$ of the square $\displaystyle ABCD$, with $\displaystyle {A}={1+i}$ and $\displaystyle {B}={4+2i}$ and the vertices anticlockwise:

• $\displaystyle {\overrightarrow{AB}}={B-A}={3+i}$.
• $\displaystyle \overrightarrow{BC}$ is $\displaystyle \overrightarrow{AB}$ turned a quarter turn: $\displaystyle {(3+i)i}={-1+3i}$.
• $\displaystyle {C}={B+(-1+3i)}={3+5i}$, and $\displaystyle {D}={A+(-1+3i)}={4i}$.

Calculate, and write the result in polar form.

Given $\displaystyle {z_1}={1+i}$ and $\displaystyle {z_2}={1+\sqrt{3}i}$.

Write $\displaystyle z_1\cdot z_2$ in polar form.

Write $\displaystyle z_1\cdot z_2$ in algebraic form.

Compare the real parts in the two forms, and find the exact value of $\displaystyle \cos\frac{7\pi}{12}$.

Find the point you get, in algebraic form.

The point $\displaystyle 3+i$ is turned about the origin by a quarter turn anticlockwise.

The point $\displaystyle 2\sqrt{3}+2i$ is turned about the origin by the angle $\displaystyle \frac{\pi}{6}$.

The point $\displaystyle 5+2i$ is turned about the point $\displaystyle 1+i$ by a quarter turn anticlockwise.

$\displaystyle ABC$ is an equilateral triangle, its vertices anticlockwise.
$\displaystyle {A}={2}$ and $\displaystyle {B}={6}$.
Find $\displaystyle C$: turn $\displaystyle B$ about $\displaystyle A$ by the angle $\displaystyle \frac{\pi}{3}$.