Question

How to solve ?please help!

Answer 1

`a=3/10` and `b=9/10`

Explanation:

As `z=a+bi`, its conjugate is `z"*"=a-bi`

and `2/z+1/(z"*")`

= `2/(a+bi)+1/(a-bi)`

= `(2a-2bi+a+bi)/(a^2+b^2)`

= `(3a)/(a^2+b^2)-b/(a^2+b^2)i`

and as `2/z+1/(z"*")=1-i`, we must have

`(3a)/(a^2+b^2)=1` and `b/(a^2+b^2)=1`

Hence `3a=b` and `a^2+b^2=a^2+9a^2=10a^2`

Hence `(3a)/(10a^2)=1` i.e. `10a=3` and `a=3/10`

and `b=9/10`

Check - We have `z=3/10+9/10i` and `z"*"=3/10-9/10i`

and `2/z+1/(z"*")=2/(3/10+9/10i)+1/(3/10+9/10i)`

= `(6/10-18/10i+3/10+9/10i)/((3/10)^2+(9/10)^2)`

= `(9/10-9/10i)/(90/100)`

= `(9/10-9/10i)/(9/10)`

= `1-i`