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

Nov 6, 2015

$\sqrt{3 + i} = \left(\sqrt{\frac{\sqrt{10} + 3}{2}}\right) + \left(\sqrt{\frac{\sqrt{10} - 3}{2}}\right) i$

#### Explanation:

Suppose ${\left(a + b i\right)}^{2} = 3 + i$

${\left(a + b i\right)}^{2} = \left({a}^{2} - {b}^{2}\right) + 2 a b i$

So equating real and imaginary parts we get:

${a}^{2} - {b}^{2} = 3$

$2 a b = 1$

Hence $b = \frac{1}{2 a}$, which we can substitute into the first equation to get:

$3 = {a}^{2} - {\left(\frac{1}{2 a}\right)}^{2} = {a}^{2} - \frac{1}{4 {a}^{2}}$

Multiply both ends by $4 {a}^{2}$ to get:

$12 \left({a}^{2}\right) = 4 {\left({a}^{2}\right)}^{2} - 1$

So:

$4 {\left({a}^{2}\right)}^{2} - 12 \left({a}^{2}\right) - 1 = 0$

From the quadratic formula we get:

${a}^{2} = \frac{12 \pm \sqrt{{12}^{2} + 16}}{8} = \frac{12 \pm \sqrt{160}}{8} = \frac{3 \pm \sqrt{10}}{2}$

Since $\sqrt{10} > 3$, choose the $+$ sign to get Real values for $a$:

$a = \pm \sqrt{\frac{\sqrt{10} + 3}{2}}$

$b = \pm \sqrt{{a}^{2} - 3} = \pm \sqrt{\frac{\sqrt{10} - 3}{2}}$

where $b$ has the same sign as $a$ since $b = \frac{1}{2 a}$

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

That is:

$\sqrt{3 + i} = \left(\sqrt{\frac{\sqrt{10} + 3}{2}}\right) + \left(\sqrt{\frac{\sqrt{10} - 3}{2}}\right) i$

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

$\sqrt{c + \mathrm{di}} = \left(\sqrt{\frac{\sqrt{{c}^{2} + {d}^{2}} + c}{2}}\right) + \left(\sqrt{\frac{\sqrt{{c}^{2} + {d}^{2}} - c}{2}}\right) i$