Question

Find the square roots of the complex number. (Enter your answers as a comma-separated list. Round terms to four decimal places.) ?

5+5i

Answer 1

`+-((sqrt((5sqrt(2)+5)/2))+(sqrt((5sqrt(2)-5)/2))i)`

`~~+-(2.4567+1.0176i)`

Explanation:

General method

Here's a non-trigonometric method for finding square roots of complex numbers...

Suppose we are given any complex number `a+bi` and want to find its square roots in the form `x+yi`.

We want to solve:

`a+bi = (x+yi)^2`

`color(white)(a+bi) = (x^2-y^2)+2xyi`

Equating real and imaginary parts, we have:

`{ (x^2-y^2=a), (2xy=b) :}`

From the second equation, we find:

`y = b/(2x)`

Substituting `b/(2x)` for `y` in the first equation, we have:

`a = x^2-(b/(2x))^2 = x^2-b^2/(4x^2)`

Multiply through by `4x^2` to get:

`4ax^2 = 4(x^2)^2-b^2`

Subtract `4ax^2` from both sides to get:

`0 = 4(x^2)^2-4a(x^2)-b^2`

`color(white)(0) = (2x^2)^2-2(2x^2)a+a^2-(a^2+b^2)`

`color(white)(0) = (2x^2-a)^2-(sqrt(a^2+b^2))^2`

`color(white)(0) = (2x^2-a-sqrt(a^2+b^2))(2x^2-a+sqrt(a^2+b^2))`

Hence:

`2x^2 = a+-sqrt(a^2+b^2)`

So:

`x^2 = (a+-sqrt(a^2+b^2))/2`

In order for `x` to be real, we need the `+` sign and hence:

`x = +-sqrt((sqrt(a^2+b^2)+a)/2)`

Then:

`y = +-sqrt(x^2-a) = +-sqrt((sqrt(a^2+b^2)-a)/2)`

Since `2xy = b`, the sign we need for `y` is the same as that for `x` if `b > 0` and the opposite of that for `x` if `b < 0`.

So if `b != 0` then the square roots of `a+bi` are:

`+-((sqrt((sqrt(a^2+b^2)+a)/2))+b/abs(b)(sqrt((sqrt(a^2+b^2)-a)/2))i)`

`color(white)()`
Example

To find the square roots of `5+5i`, put `a=5` and `b=5`.

Then:

`sqrt(a^2+b^2) = sqrt(5^2+5^2) = 5sqrt(2)`

`b/abs(b) = 5/5 = 1`

So the square roots are:

`+-((sqrt((sqrt(a^2+b^2)+a)/2))+b/abs(b)(sqrt((sqrt(a^2+b^2)-a)/2))i)`

`=+-((sqrt((5sqrt(2)+5)/2))+(sqrt((5sqrt(2)-5)/2))i)`

`~~+-(2.45673236+1.017611864i)`

`~~+-(2.4567+1.0176i)`