Square Root of Imaginary Numbers

$|3+4i|=5$ . Let the required number be $z$ . Then $|z|=\sqrt 5, arg(z)=\frac{1}{2}\tan^{-1} (\frac{4}{3})=\tan^{-1} (\frac{1}{2})$ . So

$z=\sqrt 5e^{i\tan^{-1} \frac{1}{2}}=\boxed {2+i}$ .

We are asked to find a complex number a+bi such that the square of the complex number is 4+i

So we can write this into an equation: (a+bi)^2=3+4i

Expanding we have: a^2+2abi-b^2=3+4i

So now we can make 2 equations, one with the terms with i, and one with the terms without i.

a^2-b^2=3

2abi=4i

Solving we get a=2 and b=1, so the complex number with square 3+4i is 2+i.

We need to find a complex number $z$ such that

$\begin{aligned} z^2 & = 3 + 4i & \small \blue{\text{Let }z = x+yi \text{, where }x, y \in \mathbb R} \\ (x+yi)^2 & = 3+4i \\ x^2 - y^2 + 2xyi & = 3 + 4i \end{aligned}$

Equating the real and imaginary parts $\begin{cases} x^2 - y^2 = 3 &...(1) \\ 2xy = 4 & ...(2) \end{cases}$ . From $(2): \quad y = \dfrac 2x$ . Putting it in

$\begin{aligned} (1): \quad x^2 - \frac 4{x^2} & = 3 \\ x^4 - 3x^2 - 4 & = 0 \\ (x^2 - 4)(x^2 +1 ) & = 0 & \small \blue{\text{Since }x \text{ is real}} \\ \implies x^2 & = 4 \\ x & = 2 \\ \implies y & = \frac 22 = 1 \\ \implies z & = \boxed {2+i} \end{aligned}$

\[\begin{align} (2 + i)^2 &= 2^2 + 2(2)(i) + i^2 \\ &= 4 + 4i -1 \\ &= \boxed{3 + 4i}

\end{align}\]