Find the square roots of the complex number. (Enter your answers as a comma-separated list. Round terms to four decimal places.) ?

General method

Here's a non-trigonometric method for finding square roots of complex numbers...

Suppose we are given any complex number #a+bi# and want to find its square roots in the form #x+yi#.

We want to solve:

#a+bi = (x+yi)^2#

#color(white)(a+bi) = (x^2-y^2)+2xyi#

Equating real and imaginary parts, we have:

#{ (x^2-y^2=a), (2xy=b) :}#

From the second equation, we find:

#y = b/(2x)#

Substituting #b/(2x)# for #y# in the first equation, we have:

#a = x^2-(b/(2x))^2 = x^2-b^2/(4x^2)#

Multiply through by #4x^2# to get:

#4ax^2 = 4(x^2)^2-b^2#

Subtract #4ax^2# from both sides to get:

#0 = 4(x^2)^2-4a(x^2)-b^2#

#color(white)(0) = (2x^2)^2-2(2x^2)a+a^2-(a^2+b^2)#

#color(white)(0) = (2x^2-a)^2-(sqrt(a^2+b^2))^2#

#color(white)(0) = (2x^2-a-sqrt(a^2+b^2))(2x^2-a+sqrt(a^2+b^2))#

Hence:

#2x^2 = a+-sqrt(a^2+b^2)#

So:

#x^2 = (a+-sqrt(a^2+b^2))/2#

In order for #x# to be real, we need the #+# sign and hence:

#x = +-sqrt((sqrt(a^2+b^2)+a)/2)#

Then:

#y = +-sqrt(x^2-a) = +-sqrt((sqrt(a^2+b^2)-a)/2)#

Since #2xy = b#, the sign we need for #y# is the same as that for #x# if #b > 0# and the opposite of that for #x# if #b < 0#.

So if #b != 0# then the square roots of #a+bi# are:

#+-((sqrt((sqrt(a^2+b^2)+a)/2))+b/abs(b)(sqrt((sqrt(a^2+b^2)-a)/2))i)#

#color(white)()#
Example

To find the square roots of #5+5i#, put #a=5# and #b=5#.

Then:

#sqrt(a^2+b^2) = sqrt(5^2+5^2) = 5sqrt(2)#

#b/abs(b) = 5/5 = 1#

So the square roots are:

#+-((sqrt((sqrt(a^2+b^2)+a)/2))+b/abs(b)(sqrt((sqrt(a^2+b^2)-a)/2))i)#

#=+-((sqrt((5sqrt(2)+5)/2))+(sqrt((5sqrt(2)-5)/2))i)#

#~~+-(2.45673236+1.017611864i)#

#~~+-(2.4567+1.0176i)#