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Question: \[\left( \frac{1}{1 - 2i} + \frac{3}{1 + i} \right)\left( \frac{3 + 4i}{2 - 4i} \right)\]...

(11−2i+31+i)(3+4i2−4i)\left( \frac{1}{1 - 2i} + \frac{3}{1 + i} \right)\left( \frac{3 + 4i}{2 - 4i} \right)

A

12+92i\frac{1}{2} + \frac{9}{2}i

B

12−92i\frac{1}{2} - \frac{9}{2}i

C

14−94i\frac{1}{4} - \frac{9}{4}i

D

14+94i\frac{1}{4} + \frac{9}{4}i

Answer

14+94i\frac{1}{4} + \frac{9}{4}i

Explanation

Solution

Sol. (11−2i+31+i)(3+4i2−4i)\left( \frac{1}{1 - 2i} + \frac{3}{1 + i} \right)\left( \frac{3 + 4i}{2 - 4i} \right)

=[1+2i12+22+3−3i11+12][6−16+12i+8i22+42]= \left\lbrack \frac{1 + 2i}{1^{2} + 2^{2}} + \frac{3 - 3i}{1^{1} + 1^{2}} \right\rbrack\left\lbrack \frac{6 - 16 + 12i + 8i}{2^{2} + 4^{2}} \right\rbrack

=(2+4i+15−15i10)(−1+2i2)= \left( \frac{2 + 4i + 15 - 15i}{10} \right)\left( \frac{- 1 + 2i}{2} \right)

=(17−11i)(−1+2i)20=5+45i20=14+94i.\mathbf{=}\frac{\mathbf{(17}\mathbf{-}\mathbf{11i)(}\mathbf{-}\mathbf{1 + 2i)}}{\mathbf{20}}\mathbf{=}\frac{\mathbf{5 + 45i}}{\mathbf{20}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}}\mathbf{+}\frac{\mathbf{9}}{\mathbf{4}}\mathbf{i.}