In the above example, if we let \[
z = x+iy.
\]
Then \[
z^2 = (x+iy)^2 = x^2 + 2xyi -y^2 = (x^2-y^2) + 2xyi.
\]
Thus, \[
x^2 - y^2 + 2xyi = 3+4i
\] is equivalent to \[
z^2 = 3+4i.
\]
This means that the solution of this equation gives the two distinct square roots of \(3+4i\), i.e., \(\pm sqrt(3+4i)\).
In other words, the multi-valued square-root of \(3+4i\) is given by \[
(3+4i)^{\frac{1}{2}} = \pm (2+2i).
\]

In the above example, if we let
\[
z = x+iy.
\]

Then
\[
z^2 = (x+iy)^2 = x^2 + 2xyi -y^2 = (x^2-y^2) + 2xyi.
\]

Thus,
\[
x^2 - y^2 + 2xyi = 3+4i
\]
is equivalent to
\[
z^2 = 3+4i.
\]

This means that the solution of this equation gives the two distinct square roots of \(3+4i\), i.e., \(\pm sqrt(3+4i)\).

In other words, the multi-valued square-root of \(3+4i\) is given by
\[
(3+4i)^{\frac{1}{2}} = \pm (2+2i).
\]

Remark about square roots

Remark about square roots