Sides of the Right angled Triangle in Complex notation.

Question

If $z=a+ib$ is a complex number, then $z, iz, z+iz$ represents sides of the right angled triangle.

I got this result through Cartesian form, i,e. $(a,b)$, $(-b,a)$ and $(a-b,a+b)$ are the vertices of the right angled triangle, then we can easily prove this result.

But how can we interpret this result through amplitude?

I got struck with it.

Let amplitude of $z$ is $\theta$, then amplitude of $iz$ becomes $90+\theta$, therefore it forms right angle. But How to say amplitude of $z+iz$? I got $(90+2\theta)/2$. What is the meaning? Please help me.

Answer 1

I think you mean argument but nevertheless, you have described a right angled isosceles triangle (since $z$ and $i z$ have the same magnitude) hence the other angles are each 45 degrees, which is what the arguments show.

Answer 2

$$z+iz = z(1+i) = \sqrt{2}re^{i(\theta + \pi/4)}$$ It helps to draw a picture. But in this case, since $\theta$ is just a rotation parameter common to all angles, put $\theta=0$. The first point is at $(r,0)$. Second at $(0,r)$. The third at $(r,r)$. Observe that (r,r) has length $\sqrt{2}r$ and angle 45 degrees more than the first point. Hope this clears it up.