We already have

\(LHS=\sqrt5 e^{i(\frac{\pi}{2}-x_1)},x_1=\arctan \frac12\) (1)

Continue with

\(RHS=1+2i=\sqrt5 e^{ix_2}, \ x_2=\arctan 2\) (2)

\(x_1,x_2\) are complementary angles, \(x_2=\frac{\pi}{2}-x_1 \to LHS=RHS\)

Note: polar form represented by exponential form for simplicity in (1)(2). There is no difference between two forms since r and \(\theta\) are same for LHS and RHS.