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

Aug 28, 2017

$\sqrt{3 + 7 i} {\left(12 + 5 i\right)}^{2} = a + b i$, where $a = 119 \sqrt{\frac{\sqrt{58}}{2} + \frac{3}{2}} - 120 \sqrt{\frac{\sqrt{58}}{2} - \frac{3}{2}}$ and $b = 120 \sqrt{\frac{\sqrt{58}}{2} + \frac{3}{2}} + 119 \sqrt{\frac{\sqrt{58}}{2} - \frac{3}{2}}$

#### Explanation:

Let $\sqrt{3 + 7 i} = x + i y$, then $3 + 7 i = \left({x}^{2} - {y}^{2}\right) + 2 i x y$

Therefore ${x}^{2} - {y}^{2} = 3$ and $2 x y = 7$ and

${\left({x}^{2} + {y}^{2}\right)}^{2} = {\left({x}^{2} - {y}^{2}\right)}^{2} + 4 {x}^{2} {y}^{2} = {3}^{2} + {7}^{2} = 58$

and ${x}^{2} + {y}^{2} = \sqrt{58}$

i.e. $2 {x}^{2} = \sqrt{58} + 3$ and $x = \sqrt{\frac{\sqrt{58}}{2} + \frac{3}{2}}$

and similarly $y = \sqrt{\frac{\sqrt{58}}{2} - \frac{3}{2}}$

and $\sqrt{3 + 7 i} = \sqrt{\frac{\sqrt{58}}{2} + \frac{3}{2}} + i \sqrt{\frac{\sqrt{58}}{2} - \frac{3}{2}}$

As ${\left(12 + 5 i\right)}^{2} = 144 - 25 + 120 i = 119 + 120 i$

$\sqrt{3 + 7 i} {\left(12 + 5 i\right)}^{2} = \left(119 + 120 i\right) \left(\sqrt{\frac{\sqrt{58}}{2} + \frac{3}{2}} + i \sqrt{\frac{\sqrt{58}}{2} - \frac{3}{2}}\right)$

= $119 \sqrt{\frac{\sqrt{58}}{2} + \frac{3}{2}} + 120 i \sqrt{\frac{\sqrt{58}}{2} + \frac{3}{2}} + i 119 \sqrt{\frac{\sqrt{58}}{2} - \frac{3}{2}} - 120 \sqrt{\frac{\sqrt{58}}{2} - \frac{3}{2}}$

= $\left[119 \sqrt{\frac{\sqrt{58}}{2} + \frac{3}{2}} - 120 \sqrt{\frac{\sqrt{58}}{2} - \frac{3}{2}}\right] + i \left[120 \sqrt{\frac{\sqrt{58}}{2} + \frac{3}{2}} + 119 \sqrt{\frac{\sqrt{58}}{2} - \frac{3}{2}}\right]$

Hence $\sqrt{3 + 7 i} {\left(12 + 5 i\right)}^{2} = a + b i$, where $a = 119 \sqrt{\frac{\sqrt{58}}{2} + \frac{3}{2}} - 120 \sqrt{\frac{\sqrt{58}}{2} - \frac{3}{2}}$ and $b = 120 \sqrt{\frac{\sqrt{58}}{2} + \frac{3}{2}} + 119 \sqrt{\frac{\sqrt{58}}{2} - \frac{3}{2}}$

Aug 28, 2017

$\sqrt{3 + 7 i} {\left(12 + 5 i\right)}^{2}$

$= \left(60 \sqrt{2 \sqrt{58} - 6} + \frac{119}{2} \sqrt{2 \sqrt{58} + 6}\right) + \left(60 \sqrt{2 \sqrt{58} + 6} - \frac{119}{2} \sqrt{2 \sqrt{58} - 6}\right) i$

#### Explanation:

Note that:

$\left\mid 3 + 7 i \right\mid = \sqrt{{3}^{2} + {7}^{2}} = \sqrt{9 + 49} = \sqrt{58}$

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

Note that:

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

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

Note that if:

$x = \sqrt{3 + 7 i}$

Then:

${x}^{2} = 3 + 7 i$

and:

${\left({x}^{2} - 3\right)}^{2} = {\left(7 i\right)}^{2} = - 49$

So $\sqrt{3 + 7 i}$ is the root in Q1 of:

$0 = {\left({x}^{2} - 3\right)}^{2} + 49$

$\textcolor{w h i t e}{0} = {x}^{4} - 6 {x}^{2} + 9 + 49$

$\textcolor{w h i t e}{0} = {x}^{4} - 6 {x}^{2} + 58$

Note that:

$\left({x}^{2} - k x + \sqrt{58}\right) \left({x}^{2} + k x + \sqrt{58}\right) = {x}^{4} + \left(2 \sqrt{58} - {k}^{2}\right) {x}^{2} + 58$

Equating coefficients:

$- 6 = 2 \sqrt{58} - {k}^{2}$

Hence:

${k}^{2} = 2 \sqrt{58} + 6$

So:

$k = \pm \sqrt{2 \sqrt{58} + 6}$

Thus $\sqrt{3 + 7 i}$ is the Q1 zero of:

$\left({x}^{2} - \sqrt{2 \sqrt{58} + 6} x + \sqrt{58}\right) \left({x}^{2} + \sqrt{2 \sqrt{58} + 6} x + \sqrt{58}\right)$

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 $a = 1$, $b = - \sqrt{2 \sqrt{58} + 6}$ and $c = \sqrt{58}$, we find:

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

$\textcolor{w h i t e}{x} = \frac{\sqrt{2 \sqrt{58} + 6} \pm \sqrt{\left(2 \sqrt{58} + 6\right) - 4 \sqrt{58}}}{2}$

$\textcolor{w h i t e}{x} = \frac{\sqrt{2 \sqrt{58} + 6}}{2} \pm \frac{\sqrt{2 \sqrt{58} - 6}}{2} i$

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

The other factor is a little simpler:

${\left(12 + 5 i\right)}^{2} = \left({12}^{2} - {5}^{2}\right) + 2 \left(12\right) \left(5\right) i$

$\textcolor{w h i t e}{{\left(12 + 5 i\right)}^{2}} = \left(144 - 25\right) + 120 i$

$\textcolor{w h i t e}{{\left(12 + 5 i\right)}^{2}} = 119 + 120 i$

So:

$\sqrt{3 + 7 i} {\left(12 + 5 i\right)}^{2}$

$= \left(\frac{\sqrt{2 \sqrt{58} + 6}}{2} + \frac{\sqrt{2 \sqrt{58} - 6}}{2} i\right) \left(119 + 120 i\right)$

$= \left(60 \sqrt{2 \sqrt{58} - 6} + \frac{119}{2} \sqrt{2 \sqrt{58} + 6}\right) + \left(60 \sqrt{2 \sqrt{58} + 6} - \frac{119}{2} \sqrt{2 \sqrt{58} - 6}\right) i$