Square root of a complex number.

Let $x = \sqrt{-5+12i}$ and $(a+bi)^2 = - 5 + 12i$ , where $a$ and $b$ are real, then $x = a+bi$ .

$\begin{aligned} (a+bi)^2 & = - 5 + 12i \\ a^2-b^2 + 2abi & = -5 + 12i \end{aligned}$

Equating the real and imaginary parts:

$\implies \begin{cases} a^2 - b^2 = -5 & ...(1) \\ 2ab = 12 \implies b = \dfrac 6a & ...(2) \end{cases}$

$\begin{aligned} \implies (1): \quad a^2 - \frac {36}{a^2} & = - 5 \\ a^4 + 5a^2 - 36 & = 0 \\ (a^2 - 4)(a^2 + 9) & = 0 \\ \implies a^2 & = 4 & \small \color{#3D99F6} \text{For real value }a \\ a & = 2 & \small \color{#3D99F6} \text{Taking the positive value} \\ \implies b & = \frac 62 = 3 \\ \implies x & = \boxed{2+3i} \end{aligned}$

Let $a,b$ be real numbers such that $\sqrt{-5+12i} = a+bi$ . Then

$\begin{aligned} -5 + 12i = (a^2 - b^2) + 2abi &\implies -5=a^2 - b^2 \text{ and } ab = 6 \\ &\implies \frac{36}{a^2} = b^2 = a^2+5 \\ &\implies (a^2-4)(a^2 + 9) = (a^2)^2 + 5a^2 - 36 = 0 &\implies a = \pm 2 \end{aligned}$

• If $a=2$ then $b=\frac{6}{a} = 3$ , giving $a+bi = 2+3i$
• If $a=-2$ then $b = \frac{6}{a} = -3$ , giving $a+bi = -2 - 3i$ .

We select the first one to be our principal square root, so the answer is $\boxed{2+3i}$

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Technical Note and Answer: This method isn't very robust when you really investigate it (after all, the final step required us to almost arbitrarily choose one of the two possibilities), so more generally, we define all non-integer powers of complex numbers, by $z^r = e^{r\cdot \ln(z)}$ which is well-defined once we choose a branch of the logarithm. Therefore, a more robust interpretation of the problem would require using the formula $\sqrt{z} = z^{1/2} = e^{\frac{1}{2} \ln(z)}$ and choosing a branch for the logarithm. It's then conventional to define the principal branch by defining the angle $\theta$ that $z$ makes with the positive real axis so that it satisfies $-\pi < \theta \leq \pi$ . Doing so defines the principal square root as the one with a positive real part (or along the positive imaginary axis). I.e. We once again find the answer as $\boxed{2+3i}.$