# square root of a complex number [Solved!]

**Jedothek** 22 Nov 2019, 11:24

### 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) )`

or about

`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) )`
or about
`3 + i 0.036`
which is ridiculous. What is my error ?
Thanks,
Jedothek

## Re: square root of a complex number

**Murray** 23 Nov 2019, 20:02

@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`

So your answer would have been

`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`
So your answer would have been
`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

**Jedothek** 24 Nov 2019, 11:48

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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