Find the root of the complex 3+4i

What we should know is that the root 3+4i is a complex number that looks alot like a+bi.

We can say : rt(3+4i) = a+bi (Where we dont know what a & b is..yet)

and when we square both sides (rt(3+4i))^2=(a+bi)^2 | 3+4i = (a+bi)^2

we get 3+4i = a^2+2abi-b^2

We seperate the Real and Imaginary parts to get a simultainus equation

3 = a^2-b^2

4 = 2ba

if this is solved we get a= (+-)2 and b =(+-)1

to get (+-)(2+i) <--- which is the answer