If |z|=1 then prove that (z+1)/ (z-1) is a pure imaginary number

You have to use the conjugate.

If z = a + bi
then
the conjugate of z = a - bi

Normally, the conjugate of z is written as a z with a bar above it, but we cannot do that in Yahoo!Answers.

I will use w
just remember that w is the conjugate of z

If |z|=1, then |w|=1
because
|z|= sqrt(a^2 + b^2) = |w| = 1
This will allow you to use (if you need it)
b^2 = 1 - a^2

Next trick

z-1 = a + bi - 1 = (a-1) + bi
w-1 = a - bi - 1 = (a-1) - bi
therefore
w-1 is the conjugate of z-1

When dealing with fractions in the Complex field, the easiest way out is to multiply above and below by the conjugate of the denominator

(z + 1) / (z - 1)
becomes
(z + 1)(w - 1) / (z - 1)(w - 1)

[ (a+1) + bi][(a - 1) - bi] / [(a-1) + bi)][(a-1) - bi]
Let's begin with the denominator
[(a-1) + bi)][(a-1) - bi] =
(a-1)^2 - (bi)^2 =
(a-1)^2 + b^2
And we can substitute b^2 = 1 - a^2
a^2 - 2a + 1 + 1 - a^2 =
2 - 2a
(which is a real number)

Now, the numerator:
[ (a+1) + bi][(a - 1) - bi] =
expand:
(a+1)(a-1) + (a+1)(-bi) + (bi)(a-1) - (bi)^2 =
a^2 - 1 + abi - bi + abi - bi + b^2 =
substitute b^2 = 1 - a^2, and group all "i" terms at the end
a^2 - 1 + 1 - a^2 + i(2ab - 2b) =
0 + i(2ab - 2b)

Put everything back together (numerator/denominator)

[ 0 + i(2a - 2b)] / (2 - 2a)

0 divided by anything remains 0
Leaving only
i[(2a - 2b)/(2 - 2a)]

which is a purely imaginary number
(the real part is 0)

(z+1)/(z-1)

(1+1)/(1-1)

(2)/(0)

= 0