Question

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

Answer 1

`+-sqrt5 cis(26.565^o)`, 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+4i)=(3+i4)^(1/2)` that has two values.

They are `sqrt5(cos a+isina)^(1/2)`,

where `cos a =3/5 and sin a = 4/5`

`=sqrt5(cos(a+2kpi)+isin(a+2kpi))^(1/2), k =0, 1`

`sqrt5(cos(a/2+kpi)+isin(a/2+kpi)), k =0, 1`,

using De Moivre's theorem

`=sqrt5(cos (a/2)+isin(a/2)) and sqrt5(cos (a/2+pi)+isin(a/2+pi))`

`=+-sqrt5(cos (a/2)+isin(a/2))`

`=sqrt5cis(26.565^o)`, nearly.

using `cos (pi+theta)=-cos theta and sin (pi+theta)=-sin theta`.

Answer 2

`sqrt(3+4i) = 2+i = (sqrt(5), tan^(-1)(1/2))`

Explanation:

Note that:

`(2+i)^2 = 4+4i+i^2 = 3+4i`

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

Further note that:

`abs(2+i) = sqrt(2^2+1^2) = sqrt(5)`

So we have:

`sqrt(3+4i) = 2+i = (sqrt(5), tan^(-1)(1/2))`