Hi Guest!
Let \(z=\sqrt{11+60i}\) where z is a complex number; hence of the form: \(z=x+iy\) where x and y are real numbers.
Thus, squaring both sides:
\((x+iy)^2=11+60i \\ \iff x^2-y^2+2xyi=11+60i \\ \text{Equating the real parts and imaginary parts, gives a system of two equations:} \\ x^2-y^2=11 \space \space \space \space \space \space (1)\\ 2xy=60 \implies xy=30 \space \space \space \space \space \space(2) \text{From (2), we make x or y the subject, let us make y the subject, then we substitute in (1)} \\ y=\dfrac{30}{x}. \space\\ \text{So, by (1):} \\ x^2-(\dfrac{30}{x})^2=11 \iff x^2-\frac{900}{x^2}=11 \iff x^4-900=11x^2 \iff x^4-11x^2-900=0\)
Solving for x (Use the substitution \(t=x^2\), then apply the quadratic formula) we get:
\(x_{1,2}=\pm6\)
So, \(y_{1,2}=\pm5\) (By (2)).
Therefore, the two square roots are:
\(z_1=6+5i \\ z_2=-6-5i\)
I hope this helps!
