# Thread: Square roots of complex numbers.

1. As part of a question I have to find the square root of -24 + 10i. I know that 1 + 5i is one square root, but I have no idea how to find this other than guess and check. I tried converting to polar form, but yet again it's the argument that's getting me. I know that x = -24, y = 10, and r = 26. But there doesn't seem to be a "nice" argument based on those values. arctan(10/-24) = 157.380... degrees, arccos(-24/26) = 157.380... degrees, and arcsin(10/26) = 157.380... degrees. So it's consistent, but it seems like there's no way it's going to give me an exact value for the square root, even though I know that there's a 'nice' root (1 + 5i). I thought about maybe just leaving the argument as (arccos(-24/26)) and then manipulating double angle identities, but I wasn't sure how to properly do this. Any help?

2. In this the answer is simple: $\displaystyle (-1-5i)$.
Square roots occur as negatives,

3. I guess my question wasn't clear enough. I understand that (-1-5i) would also be a square root, I just don't understand how to find +/-(1+5i) in the first place other than guess and check. The method I was taught was to convert everything into polar form, but (-24+10i) doesn't seem to convert nicely, and decimal answers won't do, I need exact answers.

4. Originally Posted by TheGreenLaser
I guess my question wasn't clear enough. I understand that (-1-5i) would also be a square root, I just don't understand how to find +/-(1+5i) in the first place other than guess and check. The method I was taught was to convert everything into polar form, but (-24+10i) doesn't seem to convert nicely, and decimal answers won't do, I need exact answers.
The easiest approach here is to let the square root be $\displaystyle a + ib$ where a and b are real. Then:

$\displaystyle -24 + 10 i = (a + ib)^2 = (a^2 - b^2) + i(2ab)$.

Equate real and imaginary parts:

$\displaystyle -24 = a^2 - b^2$ .... (1)

$\displaystyle 10 = 2ab \Rightarrow 5 = ab \Rightarrow a = \frac{5}{b}$ .... (2)

Substitute (2) into (1): $\displaystyle -24 = \frac{25}{b^2} - b^2 \Rightarrow b^4 - 24b^2 - 25 = 0 \Rightarrow (b^2 - 25)(b^2 + 1) \Rightarrow b^2 = 25$ (the other solution is rejected - why?).

etc.

5. Another approach is to remember that there are 2 square roots and they are evenly spaced around a circle. So the other square root has the same length and differs from the other by an angle of $\displaystyle \displaystyle \frac{\pi}{2}$.

6. Originally Posted by mr fantastic
The easiest approach here is to let the square root be $\displaystyle a + ib$ where a and b are real. Then:

$\displaystyle -24 + 10 i = (a + ib)^2 = (a^2 - b^2) + i(2ab)$.

Equate real and imaginary parts:

$\displaystyle -24 = a^2 - b^2$ .... (1)

$\displaystyle 10 = 2ab \Rightarrow 5 = ab \Rightarrow a = \frac{5}{b}$ .... (2)

Substitute (2) into (1): $\displaystyle -24 = \frac{25}{b^2} - b^2 \Rightarrow b^4 - 24b^2 - 25 = 0 \Rightarrow (b^2 - 25)(b^2 + 1) \Rightarrow b^2 = 25$ (the other solution is rejected - why?).

etc.
So why is the one solution rejected?
$\displaystyle (b^2 + 1) = 0 \Rightarrow b^2 = -1 \rightarrow b = i, -i$

I've substituted these values back into equation (2) to get a value for a, $\displaystyle a = \frac{5}{b} = \frac{5}{i} = \frac{5i}{i^2} = \frac{5i}{-1} = -5i$ and then used that value to form the complex number $\displaystyle a +bi = -5i + (i)i = -5i -1 = -1 - 5i$ which is one of the roots I'm looking for. Using $\displaystyle b = -i$ just yields $\displaystyle 1 + 5i$ as a root. Is it thrown out because a is supposed to represent the real part, but in that case it becomes the imaginary part? Or is it maybe because it represents the exact same roots as the other?

7. Originally Posted by TheGreenLaser
So why is the one solution rejected?
$\displaystyle (b^2 + 1) = 0 \Rightarrow b^2 = -1 \rightarrow b = i, -i$

I've substituted these values back into equation (2) to get a value for a, $\displaystyle a = \frac{5}{b} = \frac{5}{i} = \frac{5i}{i^2} = \frac{5i}{-1} = -5i$ and then used that value to form the complex number $\displaystyle a +bi = -5i + (i)i = -5i -1 = -1 - 5i$ which is one of the roots I'm looking for. Using $\displaystyle b = -i$ just yields $\displaystyle 1 + 5i$ as a root. Is it thrown out because a is supposed to represent the real part, but in that case it becomes the imaginary part? Or is it maybe because it represents the exact same roots as the other?

snip
$\displaystyle a + ib$ where a and b are real.
I will let you think about why a and b are required to be real.

$\displaystyle b = 5 \Rightarrow a = 1$, $\displaystyle b = -5 \Rightarrow a = -1$. Therefore the two roots are 1 + 5i and -1 - 5i.

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### FIND THE SQUARE ROOT OF 24 10i

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