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. ?

Question

1377 views around the world

You can reuse this answer
Creative Commons License

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#