Question

Question: Write $\mathit{z}\mathbf{=}\mathbf{-}\mathbf{1}\mathbf{+}\mathit{i}$ in polar form as $\mathit{z}\mathbf{=}\mathit{r}{\mathit{e}}^{\mathbf{i}\mathbf{\theta }}$.

Short answer

Answer

The solution is$z=\sqrt{2}{e}^{\frac{3\pi }{4}}$ .

Step-by-step solution

Given

$z=-1+i$

Explanation of the solution  

Consider the equation as follows.

$z=-1+i$

So, the following condition holds as follows.

$$\begin{gathered} z=\left({\left(1\right)}^2+{\left(-1\right)}^2\right)e^{ta{n}^{-1}\left(\frac{1}{-1}\right)i} \\ z=\left(1+1\right)e^{ta{n}^{-1}\left(-i\right)} \\ z=2e^{ta{n}^{-1}\left(-i\right)} \\ z=\sqrt{2}{e}^{\frac{3\pi }{4}} \end{gathered}$$

Thus, $z=-1+i$can be written in the polar form as$z=\sqrt{2}{e}^{\frac{3\pi }{4}}$ .