Answer
$5[\displaystyle \cos(\tan^{-1}\frac{4}{3})+i\sin(\tan^{-1}\frac{4}{3})]$
$=5[\displaystyle \cos(0.927)+i\sin(0.927)]$
Work Step by Step
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 $r(\cos{\theta}+i\sin{\theta})$:
$5[\displaystyle \cos(\tan^{-1}\frac{4}{3})+i\sin(\tan^{-1}\frac{4}{3})]$
$=5[\displaystyle \cos(0.927)+i\sin(0.927)]$