Finding real and imaginary parts of the following function

Question

I would like to find the real and imaginary parts of (1) $5^i$, (2) $i^{i+1}$ and (3) $cos(i-1)$.

Answer 1

$(1)$

$$ 5^i = e^{i\ln 5} = e^{i(\ln 5)}. $$

Hence you have $\Re(5^i) = \cos(\ln 5)$ and $\Im(5^i) = \sin(\ln 5)$. Now you could, in theory, add a period of $2kπ$ to the arguments of the sine and cosine. But that wouldn't change the value.

$(2)$

$$ i^{i+1} = i^i \cdot i = (e^{\frac{iπ}{2}})^i \cdot i = ie^{-\frac{π}{2}}. $$

Hence, $\Re(i^{i+1}) = 0$, $\Im(i^{i+1}) = e^{-\frac{π}{2}}$.

$(3)$

$$ \cos(a+ib) = \cos a \cos ib - \sin a \sin ib. $$

Now, we know that $\cosh x = \cos ix$ and $\sinh x = \sin ix$. We get

$$ \cos(a + ib) = \cos a \cosh b - \sin a \sinh b $$

As you can see, $\cos(a + ib)$ is purely real. Hence, $\Re(\cos(i-1)) = \cos 1 \cosh 1 + \sin 1 \sinh 1$ and $\Im(\cos(i-1)) = 0$.