If z=a+bi and z* is the conjugate of z. Find the value of a and b when 2/z + 1/z* = 1-i. ?

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Answer 1 · 2017-11-11T09:28:06

The answer is $a=3/10$ and $b=9/10$

I use $barz$ for the conjugate.

We need

$i^2=-1$

$(a+ib)(a-ib)=a^2-(i^2)b^2=a^2+b^2$

The complex number is $z=a+ib$

The conjugate is $barz=a-ib$

$2/z=2/(a+ib)=(2(a-ib))/((a+ib)(a-ib))=(2(a-ib))/(a^2+b^2)$

$1/barz=1/(a-ib)=(a+ib)/((a-ib)(a+ib))=(a+ib)/(a^2+b^2)$

Therefore,

$2/z+1/barz=(2(a-ib))/(a^2+b^2)+(a+ib)/(a^2+b^2)$

$=(2a-2ib+a+ib)/(a^2+b^2)$

$=(3a-ib)/(a^2+b^2)$

$=1-i$

Comparing the real parts and the imaginary parts

$(3a)/(a^2+b^2)=1$

$b/(a^2+b^2)=1$

Therefore,

$3a=b$

$(3a)/(a^2+(3a)^2)=1$

$3a=10a^2$

$10a^2-3a=0$

$a(10a-3)=0$

Therefore,

$a=0$ or $a=3/10$

Reject $a=0$

$a=3/10$ , $=>$ , $b=9/10$