Question

How do you find the square root of an imaginary number of the form a+bi?

Answer 1

Alternatively, solve without using trigonometry to find that 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)`

Explanation:

Suppose `a+bi = (c+di)^2`

How do we solve for `c` and `d`?

`(c+di)^2 = c^2+2cdi + d^2i^2 = (c^2-d^2) + (2cd)i`

So we want to solve:

`c^2-d^2 = a`

`2cd = b`

From the second of these, we find:

`d = b/(2c)`

So:

`d^2 = b^2/(4c^2)`

Substituting this in the first equation we get:

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

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

`4(c^2)^2-b^2 = 4ac^2`

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

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

From the quadratic formula, we find:

`c^2 = (4a+-sqrt((4a)^2+16b^2))/8 =(a+-sqrt(a^2+b^2))/2`

For `c` to be Real valued we require `c^2 >= 0`, hence we need to choose the root with the `+` sign...

`c^2 = (a+sqrt(a^2+b^2))/2`

Hence:

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

Then:

`d = +-sqrt(c^2-a)`

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

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

The remaining question is: What signs do we need to choose?

Since `2cd = b`:

If `b > 0` then `c` and `d` must have the same signs.

If `b < 0` then `c` and `d` must have opposite signs.

If `b=0` then `d=0`, so we don't have to worry.

If `b != 0` then we can use `b/abs(b)` as a multiplier to match the signs as we require to find that 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)`

Footnote

The question asked what is the square root of `a+bi`.

For positive Real numbers `x`, the principal square root of `x` is the positive one, which is the one we mean when we write `sqrt(x)`. This is also the square root that people commonly mean when they say "the square root of `2`" and suchlike, neglecting the fact that `2` has two square roots.

For negative Real numbers `x`, then by convention and definition, `sqrt(x) = i sqrt(-x)`, That is, the principal square root is the one with a positive coefficient of `i`.

It becomes more complicated when we deal with the square roots of Complex numbers in general.

Consider:

`sqrt(-1/2-sqrt(3)/2i) = +-(1/2-sqrt(3)/2 i)`

Which sign do you prefer?