Question #c5258

1 Answer
namralos
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*4^(-j)#.
Now #4^5 = 1024#, so we have
#1024*4^(-j)#.

Now use the trig form: #a^(jc) = cos(c*ln(a)) + jsin(c*ln(a))#.
In this case, a = 4 and c = -1.
#4^(-j) = cos(-ln(4)) + jsin(-ln(4))#
Since cosine is EVEN and sine is ODD, we have...
#4^(-j) = cos(ln(4)) - jsin(ln(4))#

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.

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