How to do this expression with complex numbers?

Question

$\frac{1}{2^{2+4i}}$, where i is the imaginary number $\sqrt{-1}$. What is the value?

Answer 1

Using Intuition behind euler's formula

$$\dfrac1{2^{2+4i}}=\dfrac{16^{-i}}{2^2}=\dfrac{(e^{\ln16})^{-i}}4 $$

$$(e^{\ln16})^{-i}=e^{-i\ln16}=\cos(-\ln16)+i\sin(-\ln16)$$

$\cos(-A)=+\cos A,\sin(-A)=-\sin A$

Now $\ln16=\ln(2^4)=4\ln2$