Here is how to find the square roots of any conplex number #a+bi# with #b# nonzero.
We seek #x+yi# such that
#(x+yi)^2=a+bi#.
Remember the product rule for complex numbers, thus
#(x+yi)^2=(x^2-y^2)+2xyi=a+bi#.
Match real and imaginary parts:
#x^2-y^2=a# Equation 1
#2xy=b# Equation 2.
We have another relation, the magnitude of #(x+yi)^2=a+bi# is the square of the magnitude of #x+yi#:
#x^2+y^2=\sqrt{a^2+b^2}# Equation 3
Now take the average of Equations 1 and 3:
#x^2={\sqrt{a^2+b^2}+a}/2#
#color(blue)(x=\pm\sqrt{{\sqrt{a^2+b^2}+a}/2})#
Now take half the difference between Equations 1 and 3:
#y^2={\sqrt{a^2+b^2}-a}/2#
#color(blue)(y=\pm\sqrt{{\sqrt{a^2+b^2}-a}/2})#
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")#
#color(blue)("x and y have opposite sign if b is negative")#
Now put this all together for the square root of #-25i#.
#a=0, b=-25#.
#b# is negative so choose opposite signs for #x# and #y#.
#\sqrt{a^2+b^2}=25#.
#x=\pm\sqrt{{\sqrt{a^2+b^2}+a}/2}=\pm\sqrt{(25+0)/2}=\pm{5\sqrt{2}}/2#.
#y=\pm\sqrt{{\sqrt{a^2+b^2}-a}/2}=\pm\sqrt{(25-0)/2}=\pm{5\sqrt{2}}/2#.
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\sqrt{2}}/2)+({5\sqrt{2}}/2)i#