Find all solutions $z^2\overline{z}^3=32$

$z^2\overline{z}^3=32$

I have bought about the following way:

1. to simplify it by $z^2\overline{z}^3=z^2\overline{z}^2\overline{z}=(z\overline{z})^2\overline{z}=|z|^4\overline{z}$

2. to replace $z$ or by $z=x+iy$ or $z=re^{i\theta}$ which should I choose? I have tried both with no success

Yes we have that

$$z^2\overline{z}^3=z^2\overline{z}^2\overline{z}=(z\overline{z})^2\overline{z}=|z|^4\overline{z}=32$$

but then $\bar z=x$ must be real and since $x^5=32 \implies x=2$ the only solution is $z=2$.

• I think the observation that $\bar z$ must be real is the key to doing this question simply and quickly – Mark Bennet Jul 27 '18 at 16:27
• @MarcBennet I agree with you! Thanks – user Jul 27 '18 at 16:29
• @gimusi Why did $\overline{z}$ have to be real? did you conclude it from the powers $|z|^4\overline{z}=2^5$? – newhere Jul 28 '18 at 18:47
• @gimusi rethinking: a product of a real number $|z|^4$ with a real to complex number $\overline{z}$ which gives a real number then $\overline{z}$ must be real – newhere Jul 28 '18 at 18:49
• @newhere Exactly! Well done. – user Jul 28 '18 at 18:56

If you use $z=re^{i\theta}$ with $r \in \mathbb R_{\ge 0}$ and $\theta \in \mathbb R$

then $z^2\overline{z}^3=32$ gives you $r^5 e^{-i\theta}=32e^{i2n\pi}$ for integer $n$

so matching coefficients gives $r=2$ and $\theta = -2n\pi$

and thus solutions are of the form $z=2e^{-i2n\pi}$ and the only one is $z=2$

Quite often it useful to see what can you say about an absolute value of a complex number and this is such a problem.

$$z^2\overline{z}^3=32\implies |z^2\overline{z}^3|=32\implies |z|^5 = 32\implies |z|=2$$

so $$z^2\overline{z}^3=32\implies |z|^4\overline{z}=32\implies \overline{z}=2\implies z=2$$

I like using

$z = re^{i\theta}; \tag 1$

we have

$\bar z = r e^{-i\theta}; \tag 2$

we are given that

$z^2 \bar z^3 = 32; \tag 3$

using (1) and (2) yields

$r^5 e^{-i\theta} = r^2 e^{2i \theta} r^3 e^{-3i \theta} = z^2 \bar z^3 = 32, \tag 4$

whence

$r^5 = \vert r \vert^5 = \vert r^5 e^{-i\theta} \vert = \vert 32 \vert = 32; \tag 5$

it follows that

$r = 2; \tag 6$

therefore, by (4),

$32 e^{-i\theta} = 32 \Longrightarrow e^{-i\theta} = 1 \Longrightarrow e^{i\theta} = 1; \tag 7$

therefore (1) becomes

$z = 2. \tag 8$

We have

$$z^2\bar z^3 = \left(z\bar z\right)^2\bar z = |z|^4\bar z = 32\Rightarrow \bar z = \alpha + 0(-j) = \alpha = z$$

hence

$$\alpha^5 = 32\Rightarrow \alpha = 2 = z$$