Question 1083651
<pre>
Let {{{sqrt(3-4i)}}}{{{""=""}}}{{{x+yi}}}, where x and y are real.

Since complex imaginary numbers are neither positive nor
negative, there is no rule about using the radical with
complex imaginary numbers as there is for real numbers.
So I suppose they want both square roots of 3-4i.

square both sides

{{{3-4i}}}{{{""=""}}}{{{x^2-y^2+2xyi}}}

Equate real parts:

{{{x^2-y^2=3}}}

Equate imaginary parts:

{{{2xy=-4}}}

Solve the system of equations:

{{{system(x^2-y^2=3,2xy=-4)}}}

Solve the second for y and substitute in the first,
and solve for the real soultions for x, then substitute
in one of the equations to get the real solutions for y.

{{{x = "" +- 2}}}

{{{y = "" +- 1}}}

At first there are four potential solutions:

2+i, 2-i, -2+i, -2-i.

However, to check for extraneous answers, we find that

by squaring each of them, we find that only 2-i and -2+i 
are solutions.

Edwin</pre>