Complex Numbers

Question

If !!|z|=1!! and !!w = (z-1)/(z+1)!!, where !!z!=1!!. Then Re(w) is?

(a) 0

(b) 1

(c) !!(-1)/(|z+1|^2)!!

(d) !!|z/(z+1)|!!

Solution

Approach 1

Assume, !!z= x+iy!!.

Given that !!|z|=1=> sqrt(x^2+y^2)=1!!

!!=> x^2+y^2=1 ....(1)!!

!!w = (z-1)/(z+1)= (x+iy-1)/(x+iy+1)= ((x-1)+iy)/((x+1)+iy)!!

Multiple and divide with !!(x+1)-iy!!, we get

!!w= (((x-1)(x+1)+y^2)+i(2xy))/((x+1)^2+y^2)!!

Re(!!w!!)= (x^2+y^2-1)/((x+1)^2+y^2)!!

From equation !!(1)!!,

Re(!!w!!)!!=0!!.

Approach 2

We can write !!z = (1+w)/(1-w)!!

Given that !!|z|=1!!, therefore

!!|(1+w)| =|1-w|!!

Assume !!w=a+ib!!

!!=> (a+1)^2+b^2= (a-1)^2+b^2!!

!!=> 4a=0 => a=0!!, mean real part of !!w!! is zero.

You can assume !!z = e^(itheta)!! and solve it.

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