Exponential complex numbers question

Question

Find all solutions of $\mathbf{e^{4z}= −3 − 3i}$

I'm having a bit of trouble with the question above. I started out with $4z= \ln(-3-3i)$ but I don't know where to go from here.

Help would be much appreciated.

Answer 1

Here are a couple of facts to bear in mind.

• The complex exponential function is periodic, with period $2\pi i$.
• $|e^{x+iy}|=|e^x||e^{iy}|=e^x\cdot1=e^x$

Now $|-3-3i|=\sqrt{3^2+3^2}=\sqrt{18},$ so $e^x=\sqrt{18}$

If we divide both sides of $e^{x+iy}$ by $e^x,$ we get $e^{iy}=-\frac12-\frac12 i$ which has the solution $y=-{\pi\over4}$.

Therefore, $z=\frac12\ln{18}+i({\pi\over4}+2n\pi), n=0,\pm 1\pm 2,\dots$

Answer 2

We have for $$z=x+iy$$so $$e^{4(x+iy)}=e^{4x}\cos(4y)+ie^{4x}\sin(4y)=-3-3i$$ so we get $$e^{4x}\cos(4y)=-3$$ and $$e^{4x}\sin(4y)=-3$$ So we have $$\tan(4y)=1$$