Answer
$-1+3i=\sqrt {10}e^{i1.89}$
Work Step by Step
$\text{Solution:}$
Step 1: Write $-1+3i$ in the Cartesian form $z=x+iy$. $$z=-1+3i$$
Step 2: So in the complex plane, the $x$-coordinate is $-1$ and the $y$-coordinate is $3$. That is $(x,y)=(-1,3)$.
Step 3: Find $r$ using the formula: $r=|z|=\sqrt {x^2+y^2}$. $$r=|z|=\sqrt {(-1)^2+3^2}=\sqrt{1+9}=\sqrt {10
}$$
Step 4: Find $\theta$, using the formula: $\theta=\tan^{-1}\left(\frac{y}{x}\right)$. Since point $(-1,3)$ is in the second quadrant ,therefore,
$$\theta=\tan^{-1}\frac{3}{-1}=\pi-\tan^{-1}(3)=1.89$$
Step 5: Write in polar form $z=re^{i\theta}$. Here $(r,\theta)=(\sqrt {10},1.89)$.Therefore, the polar form of $-1+3i$ is $$z=re^{i\theta}=\sqrt {10}e^{i1.89}$$.
