Chapter 9 - Trigonometric Identities, Models, and Complex Numbers - 9.6 Complex Numbers and De Moivre's Theorem - Exercises and Problems for Section 9.6 - Exercises and Problems - Page 392: 4

$ 2i=2e^{\frac{\pi i}{2}}$

$\text{Solution:}$ Step 1: Write $2i$ in the form $z=x+iy$. $$z=2i=0+2i$$ Step 2: So in complex plane $x$-coordinate is $0$ and $y$-coordinate is $2$. That is $(x,y)=(0,2)$. Step 3: Find $r$ using formula $r=|z|=\sqrt {x^2+y^2}$. $$r=|z|=\sqrt {0^2+2^2}=\sqrt{ 2^2}=2$$ Step 4: Find $\theta$ using formula $\theta=\tan^{-1}\frac{y}{x}$. Since the point $(0,2)$ lies on $y$-axis with $y\geq0$. $$\theta=\tan^{-1}\frac{2}{0}=\tan^{-1}(\infty)=\frac{\pi}{2}$$ Step 5: Write in polar form $z=re^{i\theta}$. Here $(r,\theta)=(2,\frac{\pi}{2})$.Therefore, polar form of $2i$ is $$ z=re^{i\theta}=2e^{\frac{\pi i}{2}}$$.