To solve z2=3+4i, write z=x+iy, where x,y in R. Then z2 = (x+iy)2 = x2 + 2xyi -y2 = (x2-y2) + 2xyi. Thus, x2 - y2 + 2xyi = 3+4i is equivalent to z2 = 3+4i. Equating real and imaginary parts gives x2-y2=3 and xy=2. Since (x2+y2)2=32+42=25, we have x2+y2=5. Hence x2=4, y2=1, with x,y of the same sign. The two square roots are z=+-(2+i). Checking, (2+i)2=4+4i-1=3+4i; the negative root has the same square.

To solve \(z^2=3+4i\), write \(z=x+iy\), where \(x,y\in\mathbb R\).

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
\[
z^2 = 3+4i.
\]

Equating real and imaginary parts gives \(x^2-y^2=3\) and \(xy=2\). Since \((x^2+y^2)^2=3^2+4^2=25\), we have \(x^2+y^2=5\). Hence \(x^2=4\), \(y^2=1\), with \(x,y\) of the same sign.

The two square roots are \[z=\pm(2+i).\] Checking, \((2+i)^2=4+4i-1=3+4i\); the negative root has the same square.

Finding Roots of Complex Numbers

Complex Numbers

International Baccalaureate Mathematics: Analysis and Approaches Higher Level

Remark about square roots

Remark about square roots