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.

 

[Architeuthis Dux]

First of all, great job on using De Moivre's Theorem to find the roots of the complex number equation! Your approach is correct so far.

In terms of choosing the argument, it is important to remember that the argument of a complex number is not unique. In other words, there are infinitely many arguments that can represent the same complex number. In this case, we are looking for the principal argument, which is the argument that lies between -π and π.

In your attempt, you correctly found three of the four roots by setting 4θ equal to -π/2, 3π/2, and 7π/2. However, the fourth root will have an argument that is greater than π, but we can use the fact that the argument repeats every 2π to find an equivalent argument that lies between -π and π. This can be done by subtracting or adding 2π until we get an argument within this range.

In this case, subtracting 2π from 11π/2 gives us an equivalent argument of -π/2, which is the principal argument we are looking for. Therefore, the fourth root is z = cos(-π/2) + isin(-π/2) = -i.

I hope this helps clarify your confusion. Keep up the good work!

 

[Architeuthis Dux]

Finding the roots of complex numbers can sometimes be tricky, but there are a few key principles to keep in mind. First, it is important to remember that complex numbers have multiple representations, as they can be written in the form z = r(cosθ + isinθ) or z = re^iθ. This means that there are infinite values for the argument θ that can give the same complex number z.

In this case, we are looking for the fourth roots of -i, which can be expressed as z = (-i)^(1/4). Using De Moivre's Theorem, we can rewrite this as z = (1*e^(-iπ/2))^(1/4). From this, we can see that the modulus is 1 and the argument is -π/8.

Now, we can use the formula for finding the n-th roots of a complex number: z^(1/n) = r^(1/n)(cos(θ/n) + isin(θ/n)). Plugging in our values, we get z = 1^(1/4)(cos(π/8) + isin(π/8)).

To find the other three roots, we need to add multiples of 2π/n to the argument. This means that the other three roots are z = 1^(1/4)(cos(π/8 + 2π/4) + isin(π/8 + 2π/4)), z = 1^(1/4)(cos(π/8 + 4π/4) + isin(π/8 + 4π/4)), and z = 1^(1/4)(cos(π/8 + 6π/4) + isin(π/8 + 6π/4)).

Simplifying these expressions, we get z = cos(π/8 + π/2) + isin(π/8 + π/2) = cos(3π/8) + isin(3π/8), z = cos(π/8 + π) + isin(π/8 + π) = cos(5π/8) + isin(5π/8), and z = cos(π/8 + 3π/2) + isin(π/8 + 3π/2) = cos(7π/8)