complex number : exam exercise I don't realize how to solve this:

$z^4\cdot{\displaystyle {\bar {z}}} + 243 = 0$

I just know this, but it doesn't help me at all.

$z\cdot{\displaystyle {\bar {z}}}=|z|^2$
You know that $z\bar{z}=|z|^2$, so then $z^3 \cdot z\cdot \bar{z}=z^3|z^2|=-243$. 

Now, $||z^3|z^2|=|z^5|=|-243|$, so $|z|=3$. 

Going back to $z^3|z^2|=-243$, we have $z^3=-27$.


  That means the three roots are $z=3e^{\frac{\pi i}{3}},3e^{\frac{3\pi i}{3}},3e^{\frac{5\pi i}{3}} $. Do you see how those three roots come from $-27$?
Let $z=r(\cos\theta+i\sin\theta),$ where $r&gt;0$ and $0\leq \theta&lt;360^{\circ}.$

Thus, $$r^4(\cos4\theta+i\sin4\theta)r(\cos(-\theta)+i\sin(-\theta))=-243$$ or
$$r^5(\cos3\theta+i\sin3\theta)=-243,$$ which gives
$$\left|r^5(\cos3\theta+i\sin3\theta)\right|=|-243|$$ or $$r^5=243$$ or
$$r=3$$ and it remains to solve:
$$\cos3\theta+i\sin3\theta=-1.$$
Can you end it now?