The square root of \(\dot z = u + i\,v\) is a complex number \(\sqrt{\dot z} = a + i\,b\) such that
\[ (a + i\,b)^2 = u + i\,v \]To find \(a\) and \(b\), we just set the real and imaginary parts of \((a + i\,b)^2\) equal to those of \(u + i\,v\):
\[ \begin{cases} a^2 - b^2 = u \\ 2ab = v \end{cases} \qquad \Rightarrow \qquad \begin{cases} a^2 - \dfrac{v^2}{4a^2} = u \\[4pt] b = \dfrac{v}{2a} \end{cases} \]Solving the second equation for \(b\), substituting into the first one and clearing the denominators, we get the system:
\[ \begin{cases} 4a^4 - v^2 = 4a^2 u \\[4pt] b = \dfrac{v}{2a} \end{cases} \]Let us rearrange the first equation of the system:
\[ 4a^4 - 4ua^2 - v^2 = 0 \]Since this is a biquadratic equation, we set \(x = a^2\), which gives the quadratic equation:
\[ 4x^2 - 4ux - v^2 = 0 \]whose solutions are:
\[ x = \frac{4u \pm \sqrt{16u^2 + 16v^2}}{8} = \frac{u \pm \sqrt{u^2 + v^2}}{2} \]Since \(x\) and \(a\) are real numbers and \(x = a^2\), only the positive solution must be considered (\(u - \sqrt{u^2 + v^2}\) is certainly negative, because \(u < \sqrt{u^2 + v^2}\)), so:
\[ x = a^2 = \frac{u + \sqrt{u^2 + v^2}}{2} \]Now we square the second equation and substitute the value found for \(a\):
\[ b^2 = \frac{v^2}{4a^2} = \frac{v^2}{2\left(u + \sqrt{u^2 + v^2}\right)} \]Rationalizing the denominator:
\[ b^2 = \frac{v^2\left(u - \sqrt{u^2 + v^2}\right)}{2\left(u + \sqrt{u^2 + v^2}\right)\left(u - \sqrt{u^2 + v^2}\right)} = \frac{v^2\left(u - \sqrt{u^2 + v^2}\right)}{2\left(u^2 - u^2 - v^2\right)} = \frac{v^2\left(u - \sqrt{u^2 + v^2}\right)}{2\left(-v^2\right)} = \frac{-u + \sqrt{u^2 + v^2}}{2} \]Taking the square root, we get the values of \(a\) and \(b\):
\[ \sqrt{\dot z} = a + i\,b = \pm \sqrt{\frac{u + \sqrt{u^2 + v^2}}{2}} \; \pm \; i \, \sqrt{\frac{-u + \sqrt{u^2 + v^2}}{2}} \]The signs of \(a\) and \(b\) must be chosen so that \(2ab = v\): the same sign if \(v > 0\), opposite signs if \(v < 0\). This gives the two roots, which are opposite to each other.
Back to Complex numbers 1, section 5.1