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Question

Mathematics Question on Complex Numbers and Quadratic Equations

If the conjugate of (x+iy)(1−2i)(x + iy) (1 - 2i) is 1+i,1 + i, then

A

x−iy=1+i1−2ix-iy = \frac {1+i}{1-2i}

B

x+iy=1−i1−2ix+iy = \frac {1-i}{1-2i}

C

x=15x= \frac {1}{5}

D

x=−15x=- \frac {1}{5}

Answer

x+iy=1−i1−2ix+iy = \frac {1-i}{1-2i}

Explanation

Solution

Let z=(x+iy)(1−2i)z = \left(x + iy\right)\left(1 - 2i\right)
=x+iy−2xi−2i2y= x + iy - 2xi - 2i^{2}y
=x+iy−2xi+2y= x + iy - 2xi + 2y
⇒z=x+2y+i(y−2x)\Rightarrow z = x + 2y + i \left(y - 2x\right)
∴zˉ=x+2y−i(y−2x)\therefore \bar{z}=x+ 2y-i \left(y-2x\right)
According to the question,
zˉ=1+i\bar{z} = 1+i
⇒x+2y−i(y−2x)=1+i\Rightarrow x + 2y - i \left(y - 2x\right) = 1 + i
On equating the real and imaginary parts from both sides, we get
x+2y=1⇔2x+4y=2...(i)x+2y=1 \Leftrightarrow 2x+4y=2\quad ...\left(i\right)
and y−2x=−1...(ii)y - 2x = - 1\quad ... \left(ii\right)
On adding Eqs. (i) and (ii), we get
5y=1⇒y=155y = 1 \Rightarrow y = \frac{1}{5}
∴\therefore From E (i), we get
2x+4(15)=22x+ 4\left(\frac{1}{5}\right) = 2
⇒2x=2−45\Rightarrow 2x=2-\frac{4}{5}
⇒2x=65\Rightarrow 2x = \frac{6}{5}
⇒x=35\Rightarrow x = \frac{3}{5}
Taking z=x+iy=1−i1−2i×1+2i1+2iz = x + iy = \frac{1-i}{1-2i}\times\frac{1+2i}{1+2i}
=1+2i−i−2i21−4i2= \frac{1+2i-i-2i^{2}}{1-4i^{2}}
z=3+i5=35+i15z = \frac{3+i}{5} = \frac{3}{5}+i \frac{1}{5}
⇒z=35+i15,\Rightarrow z = \frac{3}{5}+ i \frac{1}{5}, which is true.