How do you find the square roots of #-25i#?

1 Answer
Aug 25, 2016

The two square roots, which are negatives of each other, are:

#({5\sqrt{2}}/2)-({5\sqrt{2}}/2)i#

#(-{5\sqrt{2}}/2)+(\{5sqrt{2}}/2)i#.

Explanation:

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#