Question

What does `sqrt(3+i)` equal in a+bi form?

Answer 1

`sqrt(3+i) = (sqrt((sqrt(10)+3)/2)) + (sqrt((sqrt(10)-3)/2)) i`

Explanation:

Suppose `(a+bi)^2 = 3+i`

`(a+bi)^2 = (a^2-b^2)+2abi`

So equating real and imaginary parts we get:

`a^2-b^2 = 3`

`2ab = 1`

Hence `b = 1/(2a)`, which we can substitute into the first equation to get:

`3 = a^2-(1/(2a))^2 = a^2-1/(4a^2)`

Multiply both ends by `4a^2` to get:

`12(a^2) = 4(a^2)^2-1`

So:

`4(a^2)^2-12(a^2)-1 = 0`

From the quadratic formula we get:

`a^2 = (12+-sqrt(12^2+16))/8 = (12+-sqrt(160))/8 = (3+-sqrt(10))/2`

Since `sqrt(10) > 3`, choose the `+` sign to get Real values for `a`:

`a = +-sqrt((sqrt(10)+3)/2)`

`b = +-sqrt(a^2-3) = +-sqrt((sqrt(10)-3)/2)`

where `b` has the same sign as `a` since `b = 1/(2a)`

The principal square root is in Q1 with `a, b > 0`

That is:

`sqrt(3+i) = (sqrt((sqrt(10)+3)/2)) + (sqrt((sqrt(10)-3)/2)) i`

In fact, if `c, d > 0` then we can similarly show:

`sqrt(c+di) = (sqrt((sqrt(c^2+d^2)+c)/2)) + (sqrt((sqrt(c^2+d^2)-c)/2)) i`