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Question: If \(4x + 3\) then \(\frac{(p + i)^{2}}{2p - i} = \mu + i\lambda,\)....

If 4x+34x + 3 then (p+i)22p−i=μ+iλ,\frac{(p + i)^{2}}{2p - i} = \mu + i\lambda,.

A

μ2+λ2\mu^{2} + \lambda^{2}

B

(p2+1)24p2−1\frac{(p^{2} + 1)^{2}}{4p^{2} - 1}

C

(p2−1)24p2−1\frac{(p^{2} - 1)^{2}}{4p^{2} - 1}

D

(p2−1)24p2+1\frac{(p^{2} - 1)^{2}}{4p^{2} + 1}

Answer

(p2−1)24p2−1\frac{(p^{2} - 1)^{2}}{4p^{2} - 1}

Explanation

Solution

a+ib=[(1−i)22]100=[−2i2]100=(−i)100a + ib = \left\lbrack \frac{(1 - i)^{2}}{2} \right\rbrack^{100} = \left\lbrack \frac{- 2i}{2} \right\rbrack^{100} = ( - i)^{100}

∴a+ib=[(i)4]25=1+0i,a + ib = \lbrack(i)^{4}\rbrack^{25} = 1 + 0i,

Rationalization of denominator, we get

a=1,b=0a = 1,b = 0

z1z2=4+5i−3+2i×−3−2i−3−2i\frac{z_{1}}{z_{2}} = \frac{4 + 5i}{- 3 + 2i} \times \frac{- 3 - 2i}{- 3 - 2i}

=1−cos⁡2θ+isin⁡θ+isin⁡θcos⁡θ+isin⁡θ−isin⁡θcos⁡θ−sin⁡2θ1+cos⁡2θ−2cos⁡θ+sin⁡2θ= \frac { 1 - \cos ^ { 2 } \theta + i \sin \theta + i \sin \theta \cos \theta + i \sin \theta - i \sin \theta \cos \theta - \sin ^ { 2 } \theta } { 1 + \cos ^ { 2 } \theta - 2 \cos \theta + \sin ^ { 2 } \theta }

=1−(cos⁡2θ+sin⁡2θ)+2isin⁡θ1+(cos⁡2θ+sin⁡2θ)−2cos⁡θ= \frac { 1 - \left( \cos ^ { 2 } \theta + \sin ^ { 2 } \theta \right) + 2 i \sin \theta } { 1 + \left( \cos ^ { 2 } \theta + \sin ^ { 2 } \theta \right) - 2 \cos \theta } z1z2=−213−i(2313)=(−213,−2313)\frac{z_{1}}{z_{2}} = \frac{- 2}{13} - i\left( \frac{23}{13} \right) = \left( \frac{- 2}{13},\frac{- 23}{13} \right)

z=1+ii=−1.z = 1 + ii = \sqrt{- 1}..