Question

What does `sqrt(3+7i)*(12+5i)^2` equal in a+bi form?

Answer 1

`sqrt(3+7i)(12+5i)^2=a+bi`, where `a=119sqrt(sqrt58/2+3/2)-120sqrt(sqrt58/2-3/2)` and `b=120sqrt(sqrt58/2+3/2)+119sqrt(sqrt58/2-3/2)`

Explanation:

Let `sqrt(3+7i)=x+iy`, then `3+7i=(x^2-y^2)+2ixy`

Therefore `x^2-y^2=3` and `2xy=7` and

`(x^2+y^2)^2=(x^2-y^2)^2+4x^2y^2=3^2+7^2=58`

and `x^2+y^2=sqrt58`

i.e. `2x^2=sqrt58+3` and `x=sqrt(sqrt58/2+3/2)`

and similarly `y=sqrt(sqrt58/2-3/2)`

and `sqrt(3+7i)=sqrt(sqrt58/2+3/2)+isqrt(sqrt58/2-3/2)`

As `(12+5i)^2=144-25+120i=119+120i`

`sqrt(3+7i)(12+5i)^2=(119+120i)(sqrt(sqrt58/2+3/2)+isqrt(sqrt58/2-3/2))`

= `119sqrt(sqrt58/2+3/2)+120isqrt(sqrt58/2+3/2)+i119sqrt(sqrt58/2-3/2)-120sqrt(sqrt58/2-3/2)`

= `[119sqrt(sqrt58/2+3/2)-120sqrt(sqrt58/2-3/2)]+i[120sqrt(sqrt58/2+3/2)+119sqrt(sqrt58/2-3/2)]`

Hence `sqrt(3+7i)(12+5i)^2=a+bi`, where `a=119sqrt(sqrt58/2+3/2)-120sqrt(sqrt58/2-3/2)` and `b=120sqrt(sqrt58/2+3/2)+119sqrt(sqrt58/2-3/2)`

Answer 2

`sqrt(3+7i)(12+5i)^2`

`= (60 sqrt(2sqrt(58)-6) + 119/2 sqrt(2sqrt(58)+6))+(60 sqrt(2sqrt(58)+6) - 119/2 sqrt(2sqrt(58)-6))i`

Explanation:

Note that:

`abs(3+7i) = sqrt(3^2+7^2) = sqrt(9+49) = sqrt(58)`

Since this is not a whole number, the square root of `3+7i` has irrational real and imaginary parts. So it's not worth trying to guess.

Note that:

• `3+7i` is in Q1, so it has principal square root in Q1 and non-principal square root in Q3.

• Its complex conjugate `3-7i` is in Q4 with square roots in Q4 and Q2.

Note that if:

`x = sqrt(3+7i)`

Then:

`x^2 = 3+7i`

and:

`(x^2-3)^2 = (7i)^2 = -49`

So `sqrt(3+7i)` is the root in Q1 of:

`0 = (x^2-3)^2+49`

`color(white)(0) = x^4-6x^2+9+49`

`color(white)(0) = x^4-6x^2+58`

Note that:

`(x^2-kx+sqrt(58))(x^2+kx+sqrt(58)) = x^4+(2sqrt(58)-k^2)x^2+58`

Equating coefficients:

`-6 = 2sqrt(58)-k^2`

Hence:

`k^2 = 2sqrt(58)+6`

So:

`k = +-sqrt(2sqrt(58)+6)`

Thus `sqrt(3+7i)` is the Q1 zero of:

`(x^2-sqrt(2sqrt(58)+6)x+sqrt(58))(x^2+sqrt(2sqrt(58)+6)x+sqrt(58))`

In order for the real part to be positive, we need the first of these two quadratics to be `0`.

So using the quadratic formula with `a=1`, `b=-sqrt(2sqrt(58)+6)` and `c=sqrt(58)`, we find:

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

`color(white)(x) = (sqrt(2sqrt(58)+6)+-sqrt((2sqrt(58)+6)-4sqrt(58)))/2`

`color(white)(x) = sqrt(2sqrt(58)+6)/2+-sqrt(2sqrt(58)-6)/2i`

In order for the imaginary part to have poositive coefficient, we need the `+` sign here.

The other factor is a little simpler:

`(12+5i)^2 = (12^2-5^2)+2(12)(5)i`

`color(white)((12+5i)^2) = (144-25)+120i`

`color(white)((12+5i)^2) = 119+120i`

So:

`sqrt(3+7i)(12+5i)^2`

`= (sqrt(2sqrt(58)+6)/2+sqrt(2sqrt(58)-6)/2i)(119+120i)`

`= (60 sqrt(2sqrt(58)-6) + 119/2 sqrt(2sqrt(58)+6))+(60 sqrt(2sqrt(58)+6) - 119/2 sqrt(2sqrt(58)-6))i`