Solving $ z \cdot |z| = 4 $

Question

I have to solve $ (z^{*})^2 \cdot z^6 = 256 $ After some transformations I have: $$ |z|\cdot z = 4 \vee |z|\cdot z = -4 \vee |z|\cdot z = 4i \vee |z|\cdot z = -4i $$ How can I solve $ |z|\cdot z = 4$ ? I have no idea how can I do that...

Answer 1

HINT

We have that

$$(z^{*})^2 \cdot z^6 = 256 \iff (z^{*})^2 z^2\cdot z^4 = 256 \iff |z|^4z^4=256 \implies |z|=2$$

As an alternative by $z=re^{i\theta}$

$$(z^{*})^2 \cdot z^6 = 256 \iff r^2e^{-i2\theta}\cdot r^6e^{i6\theta}=256$$

Answer 2

$$(z^*)^2\times z^6=256$$ $$(z^*)^2\times z^2\times z^4=256$$ and we know that: $$(z^*)^2\times z^2=|z|^4$$ $$\therefore |z|^4\times z^4=256$$ $$|z|\times z=\pm4$$ which is what you got. now: $$\left|re^{i\theta}\right|\times re^{i\theta}=\pm4$$ visualising this as coordinates we get: $$r^2\cos\theta=\pm4$$ $$r^2\sin\theta=0$$

Answer 3

Addressing strictly the question asked, if $z|z|=4$, then we know that $z\not=0$, so that $|z|$ is a positive real number, and hence $z=4/|z|$ is also a positive real number. But that implies $|z|=z$, so the equation $z|z|=4$ becomes $z^2=4$, which gives us $z=\sqrt4=2$ (since we've already established that $z$ is positive).