Question

if z=1-i/1+i then Re (z) is equal to
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Answer 1

Answer:

z = (1-i)/(1+i)

z= (1-i)*(1-i)/(1+I)*(1-i)

z= (1 - I)^2/ (1^2 - i^2)

z= (1 + i^2 - 2i)/(1 - (-1))

z= (1-1-2i)/2

z= -2i/2

z= -i

hence Re(z) = 0

Answer 2

Step-by-step explanation:

Re(z) is 0.

hope you got it.