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).