MHB I guess it will be the same as i,1-i

[AkilMAI]

Complex mapping
z → f(z) =(1 + z)/(1 − z)
1.What are the images of i and 1 − i and 2.What are the images of the real and the imaginary axes?
For i we have f(i)=(1+i)/(1-i) since i(depending on the power) can be i,-i,1,-1=>0, (1+i)/(1-i),(1-i)/(1+i)
For 1 − i we have 1,-3,(2-i)/i,(2+i)/i.
Not sure about the second part..

 

[Sudharaka]

Complex mapping
z → f(z) =(1 + z)/(1 − z)
1.What are the images of i and 1 − i and 2.

Substitute \(i\), \(1 - i\) and \(2\) for z in \(f(z)=\frac{1+z}{1-z}\)

What are the images of the real and the imaginary axes?

For \(z = i\);

\[f(i)=\frac{1+i}{1-i}\]

Now you have to find the real and complex parts of \(f(i)\). Multiply both the numerator and the denominator by \(1+i\).

\[f(i)=\frac{(1+i)^2}{1-i^2}=\frac{(1+2i+i^2)}{2}\]

\[\therefore f(i) = i\]

Hence the real part of \(f(i)\) is zero and the imaginary part is 1. We usually denote this by,

\[Re[f(i)]=0\mbox{ and }Im[f(i)]=1\]

Hope you can continue with the rest of the problem.

For i we have f(i)=(1+i)/(1-i) since i(depending on the power) can be i,-i,1,-1=>0, (1+i)/(1-i),(1-i)/(1+i)

This is wrong. \(i=\sqrt{-1}\)

For 1 − i we have 1,-3,(2-i)/i,(2+i)/i.
Not sure about the second part..

...

 

[HallsofIvy]

Complex mapping
z → f(z) =(1 + z)/(1 − z)
1.What are the images of i and 1 − i and 2.What are the images of the real and the imaginary axes?
For i we have f(i)=(1+i)/(1-i) since i(depending on the power) can be i,-i,1,-1=>0, (1+i)/(1-i),(1-i)/(1+i)

What powers are you talking about? I see only the first power of i.

$f(i)= \frac{1+ i}{1- i}= \frac{1+ i}{1- i}\frac{1+ i}{1+ i}= \frac{1+ 2i+ i^2}{1+ 1}$
$= \frac{1+ 2i- 1}{2}= i$
That is the only point in the image.

For 1 − i we have 1,-3,(2-i)/i,(2+i)/i.

Again, the image contains the single point $f(1- i)= \frac{1+ (1- i)}{1- (1- i)}= \frac{2- i}{i}= 1- 2i$

Not sure about the second part..

On the real axis, z= t+ 0i. f(t)= \frac{1+ t}{1- t} where t is real. What can you say about the real and imaginary parts of that?
On the imaginary axis, z= 0+ ti. f(ti)= \frac{1+ ti}{1- ti}= \frac{1+ ti}{1- ti}\frac{1+ ti}{1+ ti}= \frac{1+ 2ti+ i^2}{1+ t^2}$
$= \frac{2ti}{1+ t^2}$. What can you say about the real and imaginary parts of that?