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I am asked to find the value of $\log z$ where $z=-1-i$. The definition of Log used here is

$\log z = \ln r + i ( \arg z)$

I know that the point lies in the third quadrant of the complex plane My question is how to find the argument of the complex number as tan inverse of $1$ is found to be $45^\circ$ .

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  • $\begingroup$ When you write Arg (sigh, now that was edited away), do you mean the principal argument? In any case, draw a figure and you will see where $-1-i$ is located, and you will most likely be able to find its argument. $\endgroup$ – mickep Jan 2 '16 at 5:57
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To find the argument, it is probably easiest to just draw it out. The point $-1-i$ is "southwest" from the origin and thus has argument $\frac{5\pi}{4}$ (or $-\frac{3\pi}{4}$).

It may help to write in polar form and you get

$z=\sqrt{2}e^{i\frac{5\pi}{4}}$

$\ln z=\ln\sqrt{2}+i\frac{5\pi}{4}$

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You have $\tan (180^\circ+\theta)=\tan(\theta)$ so the angle is $225^\circ$ or $-135^\circ$ depending on what branch of the log you use.

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