# STEP 3 2013 Q4

This question is also the third in the STEP 3 Complex Numbers Questions.

The solution to part (ii) of this question depends on the result $1 + z^2n = (1 + z^2)(1 - z^2 + z^4 - ... + z^(2n-2))$. A common factor of $1 + z^2$ is cancelled from both sides of the equation after realising that this is also a term in the product equal to $1+ z^2n$. This is fine provided that $1 + z^2$ is not 0, i.e. if z is not $+/-i$. However the next line uses the substitution $z=i$, for which the derived result is invalid! The solution acknowledges this difficulty in the first few lines but seems to disregard it in these later stages. Perhaps I am missing something or have misunderstood the solution but it doesn’t make sense to me. Sorry for the dodgy formatting and any help would be appreciated :)

### Re: Complex Numbers

There is no problem with this. You can see the initial result as an equality of polynomials, and since $1+z^2$ is a non-zero polynomial (by non-zero, I mean "not identically equal to zero"), we are allowed to cancel it. This produces another identity about polynomials, into which we can substitute any value we like, including $z=i$. Your problem seems to be that you're confusing equations with identities: if, for example, we had the equation $x^2 = x$, and we cancel the factor of $x$ from both sides, then we lose the root $x=0$. But if we take the identity $x^2(x-1) = x(x^2-x)$, and cancel a factor of $x$ from both sides, we get the identity $x(x-1) = x^2-x$, which is still true for all values of $x$, including $x=0$, despite the cancellation.

### Re: complex numbers

That explains why I was so confused - I can see the mistake in my thought process now! Thanks for the quick response.

### Thanks!

Thanks SevenOptimus!