Complex Logarithm

Relevant wiki: Euler's Formula

$\begin{aligned} \ln(3+4i) & = \ln \left(5 \color{#3D99F6}{\left(\frac 35 + \frac 45i\right)} \right) & \small \color{#3D99F6}{\text{By Euler's formula}} \\ & = \ln \left(5 \color{#3D99F6}{e^{i \tan^{-1} \frac 43}} \right) \\ & = \ln 5 + i \tan^{-1} \frac 43 \end{aligned}$

$\implies A^2 + B^2 = (\ln 5)^2 + \left(\tan^{-1} \frac 43\right)^2 \approx \boxed{3.45}$

$\ln(3 + 4i) = A + Bi \Longrightarrow e^{\ln(3 + 4i)} = e^{A + Bi} \Longrightarrow 3 + 4i = e^{A}e^{iB} = e^{A}(\cos(B) + i\sin(B))$ .

Equating respective real and imaginary coefficients gives us that $3 = e^{A}\cos(B)$ and $4 = e^{A}\sin(B)$ , and so

$3^{2} + 4^{2} = e^{2A}(\cos^{2}(B) + \sin^{2}(B)) \Longrightarrow 5^{2} = e^{2A} \Longrightarrow 2\ln(5) = 2A \Longrightarrow A = \ln(5)$ .

Thus $3 = e^{A}\cos(B) = 5\cos(B) \Longrightarrow B = \cos^{-1}\left(\dfrac{3}{5}\right)$ , and so

$A^{2} + B^{2} = (\ln(5))^{2} + \left(\cos^{-1}\left(\dfrac{3}{5}\right)\right)^{2} = \boxed{3.45}$ to 2 decimal places.