How to solve $\sqrt {35 - 5i}$ Need some hints on how to Solve $\sqrt {35 - 5i}$
Attempt.
I factorized 5 out and it became $\sqrt {5(7-i)}$
I just want to know if it can be solved further.
Thanks.
 A: You want to solve
$$(x+iy)^2=35-5i\ .$$
Working out the square and equating real and imaginary parts,
$$x^2-y^2=35\ ,\quad 2xy=-5\ .$$
Multiply the first by $4x^2$ to get
$$4x^4-4x^2y^2=140x^2$$
and substitute from the other equation,
$$4x^4-25=140x^2\ .$$
You can now solve this as a quadratic in $u=x^2$.  One of the values will have to be rejected, you then get two possible values for $x$ and two corresponding values for $y$.
A: One way is to rewrite the $35-5i$ on polar form. We have that $|35-5i| = \sqrt{35^2+5^2} = \sqrt{1250} = 25\sqrt2$. The argument you get as $\varphi=\arctan{-5/35} = -\arctan{1/7}$. So we have:
$$\sqrt{35-5i} = \sqrt{25\sqrt2e^{-i\arctan{1/7}}} = 5\sqrt{\sqrt{2}}e^{-{i\over2}\arctan{1/7}+in\pi}$$
You could also approach it by setting solving $(x+iy)^2 = 35-5i$ and identifying real and imaginary parts. This leads to a biquadratic equation, whose expression is also quite complex.
I think that the expression $\sqrt{5(7-i)}$ is as simple as it gets, but there may be reasons that you actually want it in explicit cartesian or polar form that would motivate a more complex expression.
