Complex Exponentiation

Algebra Level 2

The complex number $(-1-i)^5$ can be written as $a+bi$ , where $a$ and $b$ are real numbers. What is the value of $a+b$ ?

Details and Assumptions:

$i$ is the imaginary unit, where $i^2=-1$ .

The answer is 8.

1 solution

Arron Kau Staff
May 13, 2014

We start by writing $-1-i$ in polar form. $|-1-i| =\sqrt{(-1)^2+(-1)^2} = \sqrt{2}$ . Thus $-1 - i = \sqrt{2}\left(-\frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}}\right) = \sqrt{2}\left(\cos \left(\frac{5\pi}{4}\right) + i \sin \left(\frac{5\pi}{4}\right)\right)$ .

De Moivre's formula gives us:

$\left(\cos \left(\frac{5\pi}{4}\right) + i \sin \left(\frac{5\pi}{4}\right)\right)^5 = \cos \left(\frac{25\pi}{4}\right) + i \sin \left(\frac{25\pi}{4}\right) = \frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}}$

Thus $(-1-i)^5 = \left(\sqrt{2}\right)^5\left(\frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}}\right) = 4+4i$ . Hence $a = b = 4$ and $a + b = 8$ .

Could somebody perhaps help me out here a little, because I am having a bit of a rough time getting the signs right: So if z = (-1-i)^5, then the absolute value of z is 2^(1/2). So far so good. Now the angle between -1 and -i should be given by the arcsin(-1/-1) = arcsin(1) = pi/4, right? Then, using Euler's formula, it can be rewritten as (2^(1/2))^5 * (cos(5 pi/4) + isin(5 pi/4)). That leaves me with 4 2^(1/2) * ( -(2^(1/2) /2) - i(2^(1/2) /2) ), which when expanded is -4 -4i. This obviously left me with an answer of -8, not 8. Now I think the problem is probably with the angle, but I'm not completely sure, as the solution says the angle between -1 and -i prior to raising z to 5 is already 5 pi/4, and after raising it to 5 it becomes 25pi/4, which turns a and b positive - but as I said before I am having difficulties seeing where this comes from, as the negatives in the original arcsin(-1/-1) cancel out, making them negative again after raising to the 5th power bit. If someone could take the time to show me what I did wrong I'd be super grateful!!!!

Btw sorry for writing the equations so stocky, but I can't seem to find a better way to write them at this moment ;)

Schlemiel LR44 - 1 day, 14 hours ago

sorry not arcsin, arctan hahaha

Schlemiel LR44 - 1 day, 14 hours ago

hang on I think I've worked it out: for any coordinate, where the real part is negative I have to add pi to arctan(-1/-1) in order to get the angle spanned from (1+0i) to (-1-i), because the arctan function cannot give all the corresponding values that work for a given input (because it's a function), right? Might not be that, but hey, at least it works now...

Schlemiel LR44 - 1 day, 13 hours ago

Complex Exponentiation

The answer is 8.

1 solution

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