Answer
$z=\displaystyle \sqrt{13}(\cos[\tan^{-1}(\frac{2}{3})]+i\sin[\tan^{-1}(\frac{2}{3})])$
Work Step by Step
See p. 604,
A complex number $z=a+bi$ has the polar (or trigonometric) form
$z=r(\cos\theta+i\sin\theta)$
where $r=|z|=\sqrt{a^{2}+b^{2}}$ and $\tan\theta=b/a$.
The number $r$ is the modulus of $z$, and $\theta$ is an argument of $z$.
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$z=4+3i$
lies in Q.I.
$r=|z|=\sqrt{(3)^{2}+(2)^{2}}=\sqrt{13}$
$\tan\theta =\displaystyle \frac{2}{3}$
In Q.I, $\displaystyle \theta=\tan^{-1}(\frac{2}{3}) \approx 0.5880$
So, approximately,
$z=\sqrt{13}(\cos 0.5880+i\sin 0.5880)$
but more precisely,
$z=\displaystyle \sqrt{13}(\cos[\tan^{-1}(\frac{2}{3})]+i\sin[\tan^{-1}(\frac{2}{3})])$
