# Question #c5258

Jul 4, 2017

Use the fact that ${j}^{3} = - j$.

#### Explanation:

Let j be the imaginary unit.
Since ${j}^{3} = - j$,
we have
${4}^{5 + {j}^{3}} = {4}^{5 - j}$
By the Product Rule for Exponents, this is equal to
${4}^{5} \cdot {4}^{- j}$.
Now ${4}^{5} = 1024$, so we have
$1024 \cdot {4}^{- j}$.

Now use the trig form: ${a}^{j c} = \cos \left(c \cdot \ln \left(a\right)\right) + j \sin \left(c \cdot \ln \left(a\right)\right)$.
In this case, a = 4 and c = -1.
${4}^{- j} = \cos \left(- \ln \left(4\right)\right) + j \sin \left(- \ln \left(4\right)\right)$
Since cosine is EVEN and sine is ODD, we have...
${4}^{- j} = \cos \left(\ln \left(4\right)\right) - j \sin \left(\ln \left(4\right)\right)$

Don't forget to multiply by 1024 to obtain the final answer in the standard form. You should be able to easily identify the real and imaginary parts now.