Determining $|z-1|$ when $z=\cos\theta +i\sin\theta$ and $\theta$ is acute 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).
 A: 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).
A: Consider the isosceles triangle with vertices at $z$, $z-1$, and $0$.  The angle at $z$ is $\theta$, and if you bisect the angle then you also bisect the opposite side and create two right triangles.  The sides opposite the angles $\frac{\theta}{2}$ have length $\displaystyle{\sin\left(\frac{\theta}{2}\right)=\frac{1}{2}|z-1|}$.
You have actually already drawn the right picture for this as part of your second image, because the segment connecting $z$ to $1$ has the same length as the segment connecting $0$ to $z-1$.  Therefore in the same triangle that shows $\cos(\theta/2)=\frac{1}{2}|z+1|$ you also have $\sin(\theta/2)=\frac{1}{2}|z-1|$.
A: If $z=x+iy$, then the modulus of $z$ is given by $\sqrt{x^2+y^2}$.
So with $z=\cos\theta +i\sin\theta$, we have $z-1=(\cos\theta-1)+i\sin\theta$. So the modulus is $$\sqrt{(\cos\theta-1)^2+\sin^2\theta}=\sqrt{\cos^2-2\cos\theta+1+\sin^2\theta}=\sqrt{2-2\cos\theta}$$
since $\cos^2\theta+\sin^2\theta=1$.
Using the identity $\cos\theta=\cos^2\frac{\theta}{2}-\sin^2\frac{\theta}{2}$ and $\cos^2\frac{\theta}{2}+\sin^2\frac{\theta}{2}=1$, we then get;
$$2-2\cos\theta=4\sin^2\frac{\theta}{2}$$ and so finding the square root gives the answer.
NOTE: Write $\theta=\frac{\theta}{2}+\frac{\theta}{2}$ and use the identity $\cos (A+B)=\cos A\cos B-\sin A\sin B$ to obtain the expression  for $\cos\theta$ above
