Find all $z$ such that $e^z=6i$ Am I on the right track to solving this?
$$e^z=6i$$
Let $w=e^z$
Thus,
$$w=6i$$
$$e^w=e^{6i}$$
$$e^w=\cos(6)+i\sin(6)$$
$$\ln(e^w)=\ln(\cos(6)+i\sin(6))$$
$$w=\ln(\cos(6)+i\sin(6))$$
$$e^z=\ln(\cos(6)+i\sin(6))$$
$$\ln(e^z)=\ln(\ln(\cos(6)+i\sin(6)))$$
$$z=\ln(\ln(\cos(6)+i\sin(6)))$$
I had another method that started by taking the natural log of both sides right away, but that leads to $\arctan(6/0)$, which is undefined...
 A: $e^z=6i$.
Let $z=x+iy$.  Note that $e^z=e^x\cdot e^{iy}$
Thus
$$e^z=e^x\cdot e^{iy}=6e^{i\left(\frac{\pi}{2}+2k\pi\right)}$$
So $e^x=6$ and so $x=\ln{6}$.
So $y=\frac{\pi}{2}+2k\pi$
Therefore you have as your solutions $z=\ln{6}+i\left(\frac{\pi}{2}+2k\pi\right)$ for integer $k$.
A: Hopefully, from all of these solutions, you know how to solve this problem. Now, let's try doing it your way. You've done everything right so far:
$$z=\ln(\ln(\cos(6)+i\sin(6)))$$
By Euler's Identity, we have $\cos(6)+i\sin(6)=e^{6i}$, so clearly, taking the $\ln$ of this is just $6i$:
$$z=\ln(6i)$$
Now, if we go back to our original equation:
$$e^z=6i$$
The equation we have at the end of all of this is just taking the $\ln$ of both sides of the original equation. Basically, everything you did is all valid, but you basically return to the original equation when we're all done with simplifying everything, which is why you were off-track.
A: Suppose $z=x+iy$ and $x$ and $y$ are real.  Then
$$
6i = 6(0 + i) = e^z = e^{x+iy} = e^x e^{iy} = e^x(\cos y + i\sin y).
$$
So $e^x = 6$ and $0+1i=\cos y + i\sin y$.  Thus $\cos y=0$ and $\sin y=1$. So $y = \pi/2$ or $\pi/2+ 2\pi n$ for some integer $n$.
A: HINT:
$$6i=e^{\log(6)+i\pi/2+i2n\pi}=e^z$$
