Complex Numbers - Finding roots

[ff_yy]

Homework Statement

Solve the equation z^4= -i

Homework Equations

De Moivre's Theorem

The Attempt at a Solution

I understand how to find the roots by equating modulus and argument but I wanted to ask how do you know which arguments to take? Because I got up to

4*theta = -Pi/2, 3*Pi/2, 7*Pi/2

then I thought that I should take -5*Pi/2 because I thought that the final argument should lie between -Pi and Pi. Is that wrong? Because the answers took 11*Pi/2 instead...

I don't understand. Please help.

 

[CompuChip]

Short answer: you are right.

Long answer:

First of all, note that for any integer n,
$$e^{i \theta} = e^{i (\theta + 2 \pi)}.$$

However, it is convention to normalize angles by subtracting multiples of 2 pi. Unfortunately two different conventions are in use: some people prefer to take all angles between -pi and pi, some prefer to use angles between 0 and 2 pi. If you are using the first [second] convention and you get a negative angle [angle > pi] you can always go to the other convention by adding [subtracting] 2pi.

Now, solving z⁴ = -i you first write $z = r e^{i \theta}, i = e^{-i \pi}$. The modulus equation gives r = 1, and then you get
$$e^{4 i \theta} = e^{-i \pi / 2}$$
(I am using the convention of angles between -pi and pi here, otherwise you would get 3pi/2 on the RHS).
The solution is given by
$$4 \theta = - \pi / 2 + 2 \pi n$$
so -- dividing by 4 --
$$\theta = - \frac{\pi}{8} + n \frac{pi}{2}$$

Now all you have to do is plug in values of n to get all the inequivalent angles between -pi and pi. You will find
4theta = -5*Pi/8, -Pi/2, 3*Pi/2, 7*Pi/2

If instead, you use the convention that angles should be between 0 and pi, you have to add 2 pi to the first two (i.e. add 8pi to 4 times the angle), and you get
4theta = 3*Pi/2, 7*Pi/2, 11*Pi/2, 15*Pi/2.

The book is apparently mixing the two conventions. Of course the answers are right, and if you want you could have written down
4theta = +75*Pi/2, -33*Pi/2, 3*Pi/2, -1593*Pi/2
if you wanted.