# Complex numbers...Watch

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Thread starter 14 years ago
#1
Ok this is from prepatory questions for engineers...

Given that (1 - z) / (i + z) is real, where z = x + iy,

show that: x = y + 1

It says find the imaginary parts (I took the complex conjugate of the denominator for this) and make it equal to zero. I did this but i can't quite get the right answer.
0
14 years ago
#2
stick 1-z/i+z = q (q is some real number)

and sub x+iy for z

then rearrange and compare real and imaginary parts to find a value for q

you should see where the desired result comes from once you get to that stage.
0
14 years ago
#3
Given that (1 - z) / (i + z) is real, where z = x + iy,

show that: x = y + 1

Substituting in z and separating into real and imaginary parts:

[(1 - x) - yi] / [x + (y+1)i]

Multiply by [x - (y+1)i]:

[(1 - x) - yi][x - (y+1)i] / [x + (y+1)i][x - (y+1)i]

Simplify:

[x(1-x) - y(y+1) - (1-x)(y+1)i - xyi] / [xÂ² + yÂ² + 2y + 1]

The number will only be real if the imaginary part is 0. So considering only the imaginary part (ignoring the denominator as it's only the numerator that counts when solving a fraction equal to zero):

-(1-x)(y+1)i - xyi
= [-(y + 1 - xy - x) - xy]i
= [-y - 1 + xy + x - xy]i
= [x - y - 1]i

x - y - 1 = 0
x = y + 1

I think you probably slipped up with all the minus signs flying around. I did a few billion times
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