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Mathematics Question on Complex Numbers and Quadratic Equations

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

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

Let z=(11−2i+21+i)(3+4i2−4i)z = \left(\frac{1}{1-2i} + \frac{2}{1+i}\right) \left(\frac{3+4i}{2-4i}\right) =[1+i+3−6i(1−2i)(1+i)][3+4i2−4i]= \left[\frac{1+i+3-6i}{\left(1-2i\right)\left(1+i\right)}\right] \left[\frac{3+4i}{2-4i}\right] =[4−5i3−i][3+4i2−4i]=[32+i2−14i]= \left[\frac{4-5i}{3-i}\right]\left[\frac{3+4i}{2-4i}\right] = \left[\frac{32+i}{2-14i}\right] =32+i2−14i×2+14i2+14i=64+448i+2i−144+196= \frac{32+i}{2-14i}\times\frac{2+14i}{2+14i} = \frac{64+448i+2i-14}{4+196} =50+450i200=14+94i= \frac{50+450i}{200} = \frac{1}{4} + \frac{9}{4}i