|z| does not have to be 1 but z does have to be real

re-arrange the equation to get m = [(1+iz)/(1-iz)]^n

It is easy to sow that |1+iz| = |1-iz| so that

|1+iz|/|1-iz| = 1.

Now the mod of the right hand side of the equation = 1^n = 1

leaving |m| = 1

The problem I have so far is that nowhere have I used the fact that z is real. Now if z is real then iz and -iz are on the imaginary axis so 1+iz and 1-iz have been shfited to the right 1 unit. I am going to need more time. get back to you.

You are right I did use the restriction that z is real. Another way to look at it with a diagram to help:

1) z being real implies that iz and -iz lie on the imaginary axis with angles of pi/2 and -pi/2 respectively.
2) That is, multiplying by i and the -i rotates z through pi/2 and -pi/2
3) Adding the 1 simply shifts iz and -iz one unit to the right leaving both of them in the first quadrant forming an isosceles triangle giving |1+iz| = |1-iz|.

4) That leads us to |1+iz| / |1-iz| = 1

5) This leads to |1+iz|^n / |1-iz|^n = [|1+iz|/|1-iz|]^n

6) We now have |m| = 1^n = 1

7) So if m = a + bi then z will be real if a^2 + b^2 = 1 or:
If m lies on the unit circle a^2 + b^2 = 1 then z will be real.

8) So by choosing a and b to be on this circle you automatically generate a unique value of the argument,t, between 0 and 2pi which can be easily found by using arcsine and arccosine.
9) You can then add on your multiples of 2pi as m = e^i(2nt) = cos(2nt) + isin(2nt)
I think we are both on the same track, it' just in how you wish to express the restrictions on m = a + bi. I think by forcing a^2 + b^2 to = 1 is nice compact way to do it.
Thanks for the problem.

So far I think I have shown that |m| = 1. So if m = a + bi then a^2 + b^2 = 1.

Is that enough or am I missing some other restriction on m?

So far I have shown that |m| = 1 so if m = a + bi then a^2+b^2 = 1. Is that enough or am I missing some other restriction?

no other restriction

|z| does not have to be 1 but z does have to be real

re-arrange the equation to get m = [(1+iz)/(1-iz)]^n

It is easy to sow that |1+iz| = |1-iz| so that

|1+iz|/|1-iz| = 1.

Now the mod of the right hand side of the equation = 1^n = 1

leaving |m| = 1

The problem I have so far is that nowhere have I used the fact that z is real. Now if z is real then iz and -iz are on the imaginary axis so 1+iz and 1-iz have been shfited to the right 1 unit. I am going to need more time. get back to you.

But you actually have used the fact that z is real. If not, |1+iz|/|1-iz| would be different of 1.

Thus, the constraint |m| = 1 is valid. But where is the proof that it is sufficient ?

I made a mistake in my previous answer, so I edited it :

You are right I did use the restriction that z is real. Another way to look at it with a diagram to help:

1) z being real implies that iz and -iz lie on the imaginary axis with angles of pi/2 and -pi/2 respectively. 2) That is, multiplying by i and the -i rotates z through pi/2 and -pi/2 3) Adding the 1 simply shifts iz and -iz one unit to the right leaving both of them in the first quadrant forming an isosceles triangle giving |1+iz| = |1-iz|.

4) That leads us to |1+iz| / |1-iz| = 1

5) This leads to |1+iz|^n / |1-iz|^n = [|1+iz|/|1-iz|]^n

6) We now have |m| = 1^n = 1

7) So if m = a + bi then z will be real if a^2 + b^2 = 1 or: If m lies on the unit circle a^2 + b^2 = 1 then z will be real.

8) So by choosing a and b to be on this circle you automatically generate a unique value of the argument,t, between 0 and 2pi which can be easily found by using arcsine and arccosine. 9) You can then add on your multiples of 2pi as m = e^i(2nt) = cos(2nt) + isin(2nt) I think we are both on the same track, it' just in how you wish to express the restrictions on m = a + bi. I think by forcing a^2 + b^2 to = 1 is nice compact way to do it. Thanks for the problem.