## Calculus 8th Edition

$5[\displaystyle \cos(\tan^{-1}\frac{4}{3})+i\sin(\tan^{-1}\frac{4}{3})]$ $=5[\displaystyle \cos(0.927)+i\sin(0.927)]$
We are given: $z=3+4i$ To find $r$ of a complex number $a+bi$, we use: $\sqrt{a^2+b^2}$: $r=\sqrt{3^{2}+4^{2}}=\sqrt{9+16}=5$ To find $\theta$, we use $\tan{\theta}=\frac{b}{a}$: $\displaystyle \tan\theta=\frac{4}{3}$ And since $z$ is in the 1st quadrant: $\theta=\tan^{-1} (\displaystyle \frac{4}{3})=0.927$ To put the number in polar form, we use $r(\cos{\theta}+i\sin{\theta})$: $5[\displaystyle \cos(\tan^{-1}\frac{4}{3})+i\sin(\tan^{-1}\frac{4}{3})]$ $=5[\displaystyle \cos(0.927)+i\sin(0.927)]$