Here is how to find the square roots of any conplex number a+bia+bi with bb nonzero.
We seek x+yix+yi such that
(x+yi)^2=a+bi(x+yi)2=a+bi.
Remember the product rule for complex numbers, thus
(x+yi)^2=(x^2-y^2)+2xyi=a+bi(x+yi)2=(x2−y2)+2xyi=a+bi.
Match real and imaginary parts:
x^2-y^2=ax2−y2=a Equation 1
2xy=b2xy=b Equation 2.
We have another relation, the magnitude of (x+yi)^2=a+bi(x+yi)2=a+bi is the square of the magnitude of x+yix+yi:
x^2+y^2=\sqrt{a^2+b^2}x2+y2=√a2+b2 Equation 3
Now take the average of Equations 1 and 3:
x^2={\sqrt{a^2+b^2}+a}/2x2=√a2+b2+a2
color(blue)(x=\pm\sqrt{{\sqrt{a^2+b^2}+a}/2})x=±√√a2+b2+a2
Now take half the difference between Equations 1 and 3:
y^2={\sqrt{a^2+b^2}-a}/2y2=√a2+b2−a2
color(blue)(y=\pm\sqrt{{\sqrt{a^2+b^2}-a}/2})y=±√√a2+b2−a2
One more thing remains: choose the right signs. That is where Equation 2 comes in.
color(blue)("x and y have the same sign if b is positive")x and y have the same sign if b is positive
color(blue)("x and y have opposite sign if b is negative")x and y have opposite sign if b is negative
Now put this all together for the square root of -25i−25i.
a=0, b=-25a=0,b=−25.
bb is negative so choose opposite signs for xx and yy.
\sqrt{a^2+b^2}=25√a2+b2=25.
x=\pm\sqrt{{\sqrt{a^2+b^2}+a}/2}=\pm\sqrt{(25+0)/2}=\pm{5\sqrt{2}}/2x=±√√a2+b2+a2=±√25+02=±5√22.
y=\pm\sqrt{{\sqrt{a^2+b^2}-a}/2}=\pm\sqrt{(25-0)/2}=\pm{5\sqrt{2}}/2y=±√√a2+b2−a2=±√25−02=±5√22.
So remembering that we have opposite signs for this case we have the two square roots
({5\sqrt{2}}/2)-({5\sqrt{2}}/2)i(5√22)−(5√22)i
({-5\sqrt{2}}/2)+({5\sqrt{2}}/2)i(−5√22)+(5√22)i
