Solving for Log(z) and log(z): -9+2i

[suspenc3]

Hi, I'm kinda confused with the following, I must be looking at it wrong

Determinie all values of log(z) and Log(z):

a.) -9+2i

im confused when it comes to the angle. It is going to be in the form $$re^{i \theta}$$. where $$r= \sqrt{85}$$

Thanks.

 

[benorin]

Know that $$\log (z)= \mbox{Log} (z) +2k\pi i$$ for $$k\in\mathbb{Z}$$ is the multivalued logarithm function, and that $$\mbox{Log} (z)= \mbox{Log} |z| i\Arg (z)+2k\pi i$$ is the principal branch of the multivalued function $$\log (z)$$.

Now if $$z=re^{i\theta}$$ then $$\log (z) =\log (re^{i\theta}= \mbox{Log} (r) + i\theta + 2k\pi i$$ and hence for $$z=-9+2i$$ we have $$r=\sqrt{85}$$ and $$\theta = \tan ^{-1} \left( -\frac{9}{2}\right)$$ so that

$$\log (-9+2i) = \mbox{Log} \sqrt{85} + i \tan ^{-1} \left( -\frac{9}{2}\right) + 2k\pi i, \mbox{ for } k\in\mathbb{Z}$$

is all values of $$\log (-9+2i)$$ and $$\mbox{Log} (-9+2i)$$ is the principal value of $$\log (-9+2i)$$ (so put $$k=0$$ in the above to get

$$\mbox{Log}(-9+2i) = \mbox{Log} \sqrt{85} + i \tan ^{-1} \left( -\frac{9}{2}\right)$$.