What is the locus of z if im ((z+1)/(z-1)) = 0?

[manjuvenamma]

What is the locus of z if im ((z+1)/(z-1)) = 0?

 

[Ishuda]

What is the locus of z if im ((z+1)/(z-1)) = 0?

To clear a complex number from the denomination of a number you can multiply by the complex congugate, so what is
\(\displaystyle \frac{(z+1) (z^*-1)}{(z-1) (z^*-1)}\)
where z^* means the complex conjugate of z.

 

[pka]

What is the locus of z if im ((z+1)/(z-1)) = 0?

This post is meaningless because you did not tell us: \(\displaystyle z\to ~?\).

 

[Ishuda]

This post is meaningless because you did not tell us: \(\displaystyle z\to ~?\).

Oh?. I was pretty sure that the set of z were those z such that
Im\(\displaystyle \Biggl{(} \frac{(z+1) (z^*-1)}{(z-1) (z^*-1)} \Biggr{)}\)= 0
where Im indicates the imaginary part of.

 

[pka]

Oh?. I was pretty sure that the set of z were those z such that
Im\(\displaystyle \Biggl{(} \frac{(z+1) (z^*-1)}{(z-1) (z^*-1)} \Biggr{)}\)= 0
where Im indicates the imaginary part of.

Is the OP about limits or the imaginary part?

 

[Ishuda]

Is the OP about limits or the imaginary part?

Putting in extra spaces I thought it was
\(\displaystyle 'What\space is\space the\space locus\space of\space z\space if\space \space im\space \space ((z+1)/(z-1)) = 0?'\)
so about imaginary parts.

 

[pka]

Putting in extra spaces I thought it was
\(\displaystyle 'What\space is\space the\space locus\space of\space z\space if\space \space im\space \space ((z+1)/(z-1)) = 0?'\)
so about imaginary parts.

In that case, \(\displaystyle \dfrac{z+1}{z-1}=\dfrac{(z+1)(\overline{z}-1)}{|z-1|^2}=\dfrac{|z|^2-2i\Im{m(z)}-1}{|z-1|^2}\)

Therefore \(\displaystyle \Im m\left(\dfrac{z+1}{z-1}\right)=~?\)

 

[manjuvenamma]

To clear a complex number from the denomination of a number you can multiply by the complex congugate, so what is
\(\displaystyle \frac{(z+1) (z^*-1)}{(z-1) (z^*-1)}\)
where z^* means the complex conjugate of z.

Thanks so much, Ishuda! You are right. I meant Imaginary part only by Im. I worked out the answer and it is that z should lie on the real axis.

 

[pka]

Thanks so much, Ishuda! You are right. I meant Imaginary part only by Im. I worked out the answer and it is that z should lie on the real axis.

Correct, but note that \(\displaystyle z\ne 1\).

 

[Ishuda]

Correct, but note that \(\displaystyle z\ne 1\).

Thanks pka,
Totally slipped my mind.