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Question: Solving \(\frac{(p^{2} + 1)^{2}}{4p^{2} + 1}\), where \(z = 3 - 4i\) for x and y real, we get....

Solving (p2+1)24p2+1\frac{(p^{2} + 1)^{2}}{4p^{2} + 1}, where z=3−4iz = 3 - 4i for x and y real, we get.

A

z4−3z3+3z2+99z−95z^{4} - 3z^{3} + 3z^{2} + 99z - 95

B

z1=1−iz_{1} = 1 - i

C

z2=−2+4iz_{2} = - 2 + 4i

D

Im(z1z2z1)={Im}\left( \frac{z_{1}z_{2}}{z_{1}} \right) =

Answer

z4−3z3+3z2+99z−95z^{4} - 3z^{3} + 3z^{2} + 99z - 95

Explanation

Solution

z2=(1+i)2=1+2i+i2=1+2i−1z^{2} = (1 + i)^{2} = 1 + 2i + i^{2} = 1 + 2i - 1

Equating real and imaginary parts both sides

z2=2i.z^{2} = 2i.

z−2=12i×ii=i2i2=−i2.z^{- 2} = \frac{1}{2i} \times \frac{i}{i} = \frac{i}{2i^{2}} = - \frac{i}{2}..