# Square roots of complex numbers.

• Nov 7th 2010, 12:01 PM
TheGreenLaser
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?
• Nov 7th 2010, 01:00 PM
Plato
In this the answer is simple: $\displaystyle (-1-5i)$.
Square roots occur as negatives,
• Nov 7th 2010, 02:00 PM
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.
• Nov 7th 2010, 02:20 PM
mr fantastic
Quote:

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.
• Nov 7th 2010, 05:56 PM
Prove It
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 $\displaystyle \displaystyle \frac{\pi}{2}$.
• Nov 7th 2010, 08:20 PM
TheGreenLaser
Quote:

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?
• Nov 7th 2010, 10:34 PM
mr fantastic
Quote:

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?

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