How do i solve for z for the following equation: #z(1+i) = bar(z) + (3+2i)#. I know that conjugate of z is a-bi?

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Answer 1 · 2018-02-03T08:07:50

#z=8-3i#

Let #z=a+bi#, then #barz=a-bi#

and we can write #z(1+i)=barz+(3+2i)# as

#(a+bi)(1+i)=(a-bi)+(3+2i)#

or #a+ai+bi+bi^2=(a-bi)+(3+2i)#

or #(a-b)+(a+b)i=(a-bi)+(3+2i)#

and separating real and imaginary parts, we get

#(a-b)+(a+b)i=(a+3)+(2-b)i#

and as they have to be equal, we get

#a-b=a+3# i.e. #b=-3# and #a+b=2-b# i.e. #a=2-2b=8#

Hence #z=8-3i#