Answer
$z=\sqrt{2}(\cos 0+i\sin 0)$
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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$z=\sqrt{2}+0i$
$r=|z|=\sqrt{(\sqrt{2})^{2}+(0)^{2}}=\sqrt{2}$
$\tan\theta =0$
$\tan 0 =0$.
By symmetry (on the unit circle),
$\theta$ can also be $\pi$ (negative real axis).
$z=0-5i$
lies on the positive real axis,
so we select the appropriate argument:
$z=\sqrt{2}(\cos 0+i\sin 0)$
