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Answer:
Step-by-step explanation:
Let's start with the expression 1/(1+i). It is simplified by multiplying numerator and denominator by its conjugate.
[tex]\dfrac{1}{1+i}=\dfrac{1-i}{(1+i)(1-i)}=\dfrac{1-i}{2}[/tex]
Subtracting this from 1, as in the denominator inside parentheses, gives its conjugate.
[tex]1-\dfrac{1-i}{2}=\dfrac{1+i}{2}[/tex]
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The 5th power can be found a couple of ways. One is to use the binomial expansion. Another is to use Euler's formula. The latter can be somewhat easier.
[tex]\dfrac{1-i}{2}=\dfrac{1}{\sqrt{2}}\angle-\!45^{\circ}\\\\\left(\dfrac{1-i}{2}\right)^5=\dfrac{1}{\sqrt{2^5}}\angle(5(-45^{\circ}))=\dfrac{1}{4\sqrt{2}}(\cos(135^{\circ})+i\sin(135^{\circ}))\\\\=\dfrac{1}{8}(-1+i)[/tex]
With these parts, we can now write the first expression as ...
[tex]\displaystyle\frac{A}{i+i}\left(\frac{1-(\frac{1}{1+i})^5}{1-(\frac{1}{1+i})}\right)=A\left(\frac{1-i}{2}\right)\left(\frac{1-\frac{-1+i}{8}}{\frac{1+i}{2}}\right)=\frac{A}{8}\cdot\frac{1-i}{1+i}\cdot(9-i)\\\\=\frac{A}{8}\cdot\frac{(1-i)^2(9-i)}{1^2-i^2}=\frac{A(-2i)(9-i)}{8\cdot2}=\boxed{-A\frac{(1+9i)}{8}}[/tex]
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Using the same parts, we can simplify the second expression.
[tex]\displaystyle\frac{A}{i}\left(1-\left(\frac{1}{1+i}\right)^5\right)=\frac{A}{i}\cdot\frac{(9-i)}{8}=\frac{A(9-i)(i)}{8i^2}=\boxed{-A\frac{(1+9i)}{8}}[/tex]
