Question

How you calculate a It is possible to calculate `sqrt(2+3i)`? And if it is,how to calculate it?

Answer 1

`sqrt(2+3i) = 1/2sqrt(2sqrt(13)+4)+1/2sqrt(2sqrt(13)-4)i`

Explanation:

Note that `2+3i` is in Q1. So its principal square root will be in Q1 too.

We could express it using trigonometric functions, but let's try an algebraic approach...

Note that:

`((sqrt(2+3i))^2-2)^2 + 9 = 0`

That is, `sqrt(2+3i)` is one of the four roots of:

`(x^2-2)^2+9 = 0`

Multiplied out this becomes:

`x^4-4x^2+13 = 0`

Note that:

`(a^2-kab+b^2)(a^2+kab+b^2) = a^4+(2-k^2)a^2b^2+b^4`

Putting `a=x` and `b=root(4)(13)`, we have:

`(x^2-kroot(4)(13)x+sqrt(13))(x^2+kroot(4)(13)x+sqrt(13)) = x^4+(2-k^2)sqrt(13)x^2+13`

We want:

`(2-k^2)sqrt(13) = -4`

So:

`2sqrt(13)+4 = sqrt(13)k^2`

So:

`k^2 = (2sqrt(13)+4)/sqrt(13) = 2+(4sqrt(13))/13`

Let:

`k = sqrt(2+(4sqrt(13))/13)`

So:

`0 = x^4-4x+13`

`color(white)(0) = (x^2-sqrt(2+(4sqrt(13))/13)root(4)(13)x+sqrt(13))(x^2+sqrt(2+(4sqrt(13))/13)root(4)(13)x+sqrt(13))`

Hence, using the quadratic formula and picking the root in Q1, we find:

`sqrt(2+3i) = 1/2sqrt(2+4sqrt(13)/13)root(4)(13)+1/2sqrt((2+4sqrt(13)/13)sqrt(13)-4sqrt(13))`

`color(white)(sqrt(2+3i)) = 1/2sqrt(2sqrt(13)+4)+1/2sqrt(2sqrt(13)-4)i`