Question

How do you multiply complex numbers in trigonometry?

Answer 1

Step 1. Formation of the trigonometric form of complex numbers.

$\mathit{z}\mathbf{=}\mathit{r}\mathbf{cos}\left(\theta \right)\mathbf{+}\mathit{i}\mathit{r}\mathbf{sin}\left(\theta \right)$

$z=r\left(\cos \left(\theta \right)+i\sin \left(\theta \right)\right)$

Where, $r$$=$magnitude or modulus and $\theta =$argumnet of $z$

$$\begin{gathered} \mathit{r}\mathbf{=}\left|z\right| \\ \mathbf{}\mathbf{=}\sqrt{\left(x^2+y^2\right)} \end{gathered}$$

$$\begin{gathered} x=r\cos \left(\mathrm{\theta }\right) \\ y=r\sin \left(\mathrm{\theta }\right) \end{gathered}$$

Step 2. Finding the multiplication of complex numbers in polar form:

If $z_1=r_1(\cos \left(\mathrm{\theta }\right)+\mathrm{isin}\left(\mathrm{\theta }\right))$ and

$z_2=r_2(\cos \left(\mathrm{\phi }\right)+\mathrm{isin}\left(\mathrm{\phi }\right))$ are two complex numbers in polar form. Then the multiplication of the two complex numbers can be determined by:

The multiplication of $z_1$ and $z_2$ is

$z_1{z}_2=$$r_1\left(\cos \left(\theta \right)+i\sin \left(\theta \right)\right)\times r_2\left(\cos \left(\phi \right)+i\sin \left(\phi \right)\right)$

$\Rightarrow$$z_1{z}_2=r_1{r}_2$$\left(\cos \left(\theta \right)+i\sin \left(\theta \right)\right)\times \left(\cos \left(\phi \right)+i\sin \left(\phi \right)\right)$

$\Rightarrow$$z_1{z}_2=r_1{r}_2\left(\cos \left(\theta \right)\cos \left(\phi \right)+i\cos \left(\theta \right)\sin \left(\phi \right)+i\sin \left(\theta \right)\cos \left(\phi \right)+i^2\sin \left(\theta \right)\sin \left(\phi \right)\right)$

$\Rightarrow$$z_1{z}_2=r_1{r}_2\left(\cos \left(\theta \right)\cos \left(\phi \right)+i\cos \left(\theta \right)\sin \left(\phi \right)+i\sin \left(\theta \right)\cos \left(\phi \right)-\sin \left(\theta \right)\sin \left(\phi \right)\right)$ ($i^2=-1$)

$\Rightarrow$$z_1{z}_2=r_1{r}_2\left(\cos \left(\theta \right)\cos \left(\phi \right)-\sin \left(\theta \right)\sin \left(\phi \right)+i\left(\cos \left(\theta \right)\sin \left(\phi \right)+\sin \left(\theta \right)\cos \left(\phi \right)\right)\right)$

$\Rightarrow$$z_1{z}_2=r_1{r}_2\left[\left(\cos \left(\theta \right)\cos \left(\phi \right)-\sin \left(\theta \right)\sin \left(\phi \right)\right)+i\left(\cos \left(\theta \right)\sin \left(\phi \right)+\sin \left(\theta \right)\cos \left(\phi \right)\right)\right]$ Using $$\begin{gathered} \left[\cos (A+B)=\cos \left(A\right)\cos \left(B\right)-\sin \left(A\right)\sin \left(B\right) \\ \sin (A+B)=\sin \left(A\right)\cos \left(B\right)+\cos \left(A\right)\sin \left(B\right)\right] \end{gathered}$$

$\Rightarrow$$z_1{z}_2={r_{1}r}_2\left[\left(\cos \left(\theta \right)\cos \left(\phi \right)-\sin \left(\theta \right)\sin \left(\phi \right)\right)\right]+i\left[\left(\cos \left(\theta \right)\sin \left(\phi \right)+\sin \left(\theta \right)\cos \left(\phi \right)\right)\right]$

$\Rightarrow$${\mathit{z}}_{\mathbf{1}}{\mathit{z}}_{\mathbf{2}}\mathbf{=}{\mathit{r}}_{\mathbf{1}}{\mathit{r}}_{\mathbf{2}}\mathbf{(}\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{(}\mathbf{\theta }\mathbf{+}\mathbf{\phi }\mathbf{)}\mathbf{+}\mathbf{i}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{(}\mathbf{\theta }\mathbf{+}\mathbf{\phi }\mathbf{)}\mathbf{)}$