# How do you convert sqrt(3+4i)  to polar form?

Jan 12, 2017

$\pm \sqrt{5} c i s \left({26.565}^{o}\right)$, nearly. See explanation.

#### Explanation:

Only for real ${a}^{2}$, sqrta^2)=|a| #, by convention ( or definition ).

So, the other square root $- | a |$ is out of reach.

Here, for sqrt(z), where z is complex, we cannot conveniently take

one and keep off the other.

So, $\sqrt{3 + 4 i} = {\left(3 + i 4\right)}^{\frac{1}{2}}$ that has two values.

They are $\sqrt{5} {\left(\cos a + i \sin a\right)}^{\frac{1}{2}}$,

where $\cos a = \frac{3}{5} \mathmr{and} \sin a = \frac{4}{5}$

$= \sqrt{5} {\left(\cos \left(a + 2 k \pi\right) + i \sin \left(a + 2 k \pi\right)\right)}^{\frac{1}{2}} , k = 0 , 1$

$\sqrt{5} \left(\cos \left(\frac{a}{2} + k \pi\right) + i \sin \left(\frac{a}{2} + k \pi\right)\right) , k = 0 , 1$,

using De Moivre's theorem

$= \sqrt{5} \left(\cos \left(\frac{a}{2}\right) + i \sin \left(\frac{a}{2}\right)\right) \mathmr{and} \sqrt{5} \left(\cos \left(\frac{a}{2} + \pi\right) + i \sin \left(\frac{a}{2} + \pi\right)\right)$

$= \pm \sqrt{5} \left(\cos \left(\frac{a}{2}\right) + i \sin \left(\frac{a}{2}\right)\right)$

$= \sqrt{5} c i s \left({26.565}^{o}\right)$, nearly.

using $\cos \left(\pi + \theta\right) = - \cos \theta \mathmr{and} \sin \left(\pi + \theta\right) = - \sin \theta$.

Mar 12, 2017

$\sqrt{3 + 4 i} = 2 + i = \left(\sqrt{5} , {\tan}^{- 1} \left(\frac{1}{2}\right)\right)$

#### Explanation:

Note that:

${\left(2 + i\right)}^{2} = 4 + 4 i + {i}^{2} = 3 + 4 i$

Since $3 + 4 i$ is in Q1, $2 + i$ is its principal square root.

Further note that:

$\left\mid 2 + i \right\mid = \sqrt{{2}^{2} + {1}^{2}} = \sqrt{5}$

So we have:

$\sqrt{3 + 4 i} = 2 + i = \left(\sqrt{5} , {\tan}^{- 1} \left(\frac{1}{2}\right)\right)$