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# square root of a complex number [Solved!]

### My question

I cant get a formula for the square root of a + bi to work.

### Relevant page

http://stanleyrabinowitz.com/bibliography/complexSquareRoot.pdf

### What I've done so far

I start with sqrt(a+ b i), translate into polar or exponential (re^(i θ)) coordinates and back again, and get

sqrt(a+b i) = sqrt(r) (cos [theta] + i sin [theta]).

where

theta = 0.5 arcsin (b/r)

r =sqrt( (a^2 +b^2)

Then I test this with (2 + 3 i )^2 = -5 + 12 i.

I get

theta = 0.5 arcsin(12/13) = 0.5880

sqrt(-5 + 12 i)  = sqrt(13)( cos[0.5880 ] + i sin [0.5880]).

which turns out to be

 sqrt( 13) ( (.83)+ i (.01) )

3 + i 0.036

which is ridiculous. What is my error ?

Thanks,

Jedothek

X

I cant get a formula for the square root of a + bi to work.
Relevant page

<a href="http://stanleyrabinowitz.com/bibliography/complexSquareRoot.pdf">http://stanleyrabinowitz.com/bibliography/complexSquareRoot.pdf</a>

What I've done so far

I start with  sqrt(a+ b i), translate into polar or exponential (re^(i θ)) coordinates and back again, and get

sqrt(a+b i) = sqrt(r) (cos [theta] + i  sin [theta]).

where

theta = 0.5 arcsin (b/r)

r =sqrt( (a^2 +b^2)

Then I test   this with (2 + 3 i )^2 = -5 + 12 i.

I get

theta = 0.5 arcsin(12/13) = 0.5880

sqrt(-5 + 12 i)  = sqrt(13)( cos[0.5880 ] + i sin [0.5880]).

which turns out to be

sqrt( 13) ( (.83)+ i (.01) )

3 + i 0.036

which is ridiculous. What is my error ?

Thanks,

Jedothek

## Re: square root of a complex number

@Jedothek: I formatted your question so it was easier to read. I encourage you to make use of the "add math" feature in this forum. (You can click "Show code" to see how I did it.)

(1) I agree with the part where you have:

cos(0.5880) = 0.83

However,

sin(0.5880) = 0.5547

sqrt(13)(0.83 + 0.5547i)  = 3 + 2i

(2) Now, the fact our numbers are the wrong way round gives a clue to where the solution went haywire.

The angle representing the complex number -5+12i is in the second quadrant, so it will be an angle between pi/2~~1.5708 and pi~~3.1416, and 0.5880 is not in this range.

So we need to make use of the Reference angle (about half-way down that page).

Our angle should be

theta = 0.5(pi - arcsin(12/13))  = 0.5xx1.9656  = 0.9828

So now we'll have

sqrt(13)( cos[0.9828] + i sin [0.9828])  = 2+3i

which is what we were hoping for.

Hope it helps.

X

@Jedothek: I formatted your question so it was easier to read. I encourage you to make use of the "add math" feature in this forum. (You can click "Show code" to see how I did it.)

(1) I agree with the part where you have:

cos(0.5880) = 0.83

However,

sin(0.5880) = 0.5547

sqrt(13)(0.83 + 0.5547i)  = 3 + 2i

(2) Now, the fact our numbers are the wrong way round gives a clue to where the solution went haywire.

The angle representing the complex number -5+12i is in the <strong>second quadrant</strong>, so it will be an angle between pi/2~~1.5708 and pi~~3.1416, and 0.5880 is not in this range.

So we need to make use of the <a href="https://www.intmath.com/trigonometric-functions/6-trigonometry-functions-any-angle.php">Reference angle</a> (about half-way down that page).

Our angle should be

theta = 0.5(pi - arcsin(12/13))  = 0.5xx1.9656  = 0.9828

So now we'll have

sqrt(13)( cos[0.9828] + i sin [0.9828])  = 2+3i

which is what we were hoping for.

Hope it helps.

## Re: square root of a complex number

Thanks so much! Here I was worrying that mathematics didn't make sense. You have restored my faith.
As I'm sure you realized, I must have been taking that sin .588 in degrees.

X

Thanks so much! Here  I  was worrying that mathematics didn't make sense. You have restored  my faith.
As I'm sure you realized, I must have  been taking that sin .588 in degrees.

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