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Question: \[\frac{1 + 7i}{(2 - i)^{2}} =\]...

1+7i(2−i)2=\frac{1 + 7i}{(2 - i)^{2}} =

A

2(cos⁡3π4+isin⁡3π4)\sqrt{2}\left( \cos\frac{3\pi}{4} + i\sin\frac{3\pi}{4} \right)

B

2(cos⁡π4+isin⁡π4)\sqrt{2}\left( \cos\frac{\pi}{4} + i\sin\frac{\pi}{4} \right)

C

(cos⁡3π4+isin⁡3π4)\left( \cos\frac{3\pi}{4} + i\sin\frac{3\pi}{4} \right)

D

None of these

Answer

2(cos⁡3π4+isin⁡3π4)\sqrt{2}\left( \cos\frac{3\pi}{4} + i\sin\frac{3\pi}{4} \right)

Explanation

Solution

Sol. 1+7i(2−i)2=(1+7i)(3+4i)(3−4i)(3+4i)=−25+25i25=−1+i\frac{1 + 7i}{(2 - i)^{2}} = \frac{(1 + 7i)(3 + 4i)}{(3 - 4i)(3 + 4i)} = \frac{- 25 + 25i}{25} = - 1 + i

Let z=x+iy=−1+iz = x + iy = - 1 + i, ∴ x=rcos⁡θ=−1x = r\cos\theta = - 1 and y=rsin⁡θ=1y = r\sin\theta = 1

∴ θ=3π4\theta = \frac{3\pi}{4} and r=2r = \sqrt{2}, Thus 1+7i(2−i)2=2[cos⁡3π4+isin⁡3π4]\frac{1 + 7i}{(2 - i)^{2}} = \sqrt{2}\left\lbrack \cos\frac{3\pi}{4} + i\sin\frac{3\pi}{4} \right\rbrackAlternative method: ∣1+7i(2−i)2∣=∣1+7i3−4i∣=2\left| \frac{1 + 7i}{(2 - i)^{2}} \right| = \left| \frac{1 + 7i}{3 - 4i} \right| = \sqrt{2} and arg(1+7i3−4i)=tan⁡−17−tan⁡−1(−43)\left( \frac{1 + 7i}{3 - 4i} \right) = \tan^{- 1}7 - \tan^{- 1}\left( \frac{- 4}{3} \right) =tan⁡−17+tan⁡−143=3π4= \tan^{- 1}7 + \tan^{- 1}\frac{4}{3} = \frac{3\pi}{4}

∴ 1+7i(2−i)2=2(cos⁡3π4+isin⁡3π4)\frac{\mathbf{1 + 7i}}{\mathbf{(2}\mathbf{-}\mathbf{i}\mathbf{)}^{\mathbf{2}}}\mathbf{=}\sqrt{\mathbf{2}}\left( \mathbf{\cos}\frac{\mathbf{3\pi}}{\mathbf{4}}\mathbf{+ i}\mathbf{\sin}\frac{\mathbf{3\pi}}{\mathbf{4}} \right)