As the question indicates we are supposed to find the modulus of z-1.
When trying to solve the problem I drew a diagram which you can see below:
The book I am working in solved a similar problem when showing examples, however the question was for $|z+1|$. There they proved that the modulus is given by $2\cos {\theta \over 2}$. In the case of $z-1$ though the shape is not a rhombus (which it is when you construct the diagram for $|z+1|$). As a result the diagonals are not perpendicular. The book however states that $|z-1|=2\sin {\theta \over 2}$
Proof that $|z+1|=2\cos {\theta \over 2}$
Using the diagram above one can see that $|z+1|=2\cos {\theta \over 2}$ as the length from the origin to the intersection of the diagonals is given by $\cos {\theta \over 2}$. Now, to get the length of the whole diagonal just multiply that by 2.
Does anyone know whether the book is right and I am missing something or if the answer is in fact different (in which case the correct answer would be appreciated).
Best Answer
You certainly can figure out the modulus using a diagram, but personally I find this method unpleasant, so I shall proceed by algebraic means. Now, the modulus of $z = x+iy$ can be defined as $|z| = \sqrt{x^2+y^2}$. For $z = \cos\theta+i\sin\theta$, $$\begin{eqnarray*}|z-1| &=& \sqrt{(\cos\theta-1)^2+\sin^2\theta}\\ &=& \sqrt{\cos^2\theta-2\cos\theta+1+\sin^2\theta}\\ &=& \sqrt{2-2\cos\theta}\\ &=& 2\sqrt{\frac{1-\cos\theta}{2}}\\ &=& 2\sin\frac{\theta}{2}\end{eqnarray*}$$ with the last equality holding only for acute angles $\theta$ (this is the so-called half-angle identity).