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z/(z-i) problem
By Patrick Aouad (P2687) on Tuesday, September 12, 2000 - 07:38 am:

Hi, I am studying complex numbers and I found this little question:

z is an element of C such that z/(z-i) is real. Show that z is imaginary.

Now I think I have done it, but I am curious as to how somebody else would do it (mainly so I can check my reasoning).

Thanks.
PAT.

By Marion Palles-Clark (Mhp21) on Tuesday, September 12, 2000 - 11:51 am:

Hi there!

OK, we have z in C such that z/(z-i) is real. So let's write
z = a(z-i) (where a is real, and I've multiplied each side by z-i)

re-arranging this gives us:
z(1-a) = ai

now we know that a is real, and so, obviously, is a-1. So the RHS is pure imaginary, and the LHS is a (real) scalar multiple of z, so z must be pure imaginary too.

If this is the way that you did it, and are still wondering about that argument, remember that we can look at the real and imaginary parts of an equation separately.

If we have a + bi = c + di (a,b,c,d all real) then we know that a = c and b = d. If, in the equation z(1-a) = ai we express z as z = x + iy (x, y in R), then we get
(1-a)x + (1-a)iy = ai.

Equating real and imaginary parts gives x = 0 and y = a/(1-a). So z = ai/(1-a) which is pure imaginary.

That was a bit long winded, but I hope it has sorted everything out!

Marion

By Patrick Aouad (P2687) on Wednesday, September 13, 2000 - 03:47 am:

Thanks Marion