a) Describe the locus defined by Arg(z-i)=Arg(z-1)
b) Describe the locus defined by Arg(z-2)=2Arg(z+1)
Firstly, we need to establish some restrictions on \(z\) to determine its argument. Note that \(\tan^{-1}{\left(\alpha\right)}\) has domain \(\alpha\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)\), thus, it gives us angles in quadrants 1 and 4, where \(x>0\).
If \(x<0\) and \(y>0\), our complex number is in quadrant 2, but \(\tan^{-1}{\left(\frac{y}{x}\right)}\) gives a value in quadrant 4. Thus, if the complex number is in the second quadrant, it's argument can be found by adding \(\pi\) to the previous expression, i.e. \(\pi+\tan^{-1}{\left(\frac{y}{x}\right)}\).
If \(x<0\) and \(y<0\), our complex number is in quadrant 3, but \(\tan^{-1}{\left(\frac{y}{x}\right)}\) gives a value in quadrant 1. Thus, if the complex number is in the third quadrant, it's argument can be found by adding \(\pi\) to the previous expression, i.e. \(\pi+\tan^{-1}{\left(\frac{y}{x}\right)}\) (note that I have, for simplicity's sake in terms of expressing the argument, not considered the principal argument here (i.e. I will get quadrant 3 angles that are between \(\pi\) and \(\frac{3\pi}{2}\)).
If \(x=0\) the value of the argument depends on the value of \(y\). That is, if \(y\) is positive, the argument is \(\frac{\pi}{2}\). If \(y\) is negative, the argument is \(-\frac{\pi}{2}\). If \(y=0\), the argument is undefined.
Now we can put it all together:
&=\tan^{-1}{\left(\frac{y}{x}\right)},x>0\\\text{Arg}\left(z\right)&=\pi+\tan^{-1}{\left(\frac{y}{x}\right)},x<0\\\text{Arg}\left(z\right)&=\frac{\pi}{2},x=0,y>0\\\text{Arg}\left(z\right)&=-\frac{\pi}{2},x=0,y<0\\\text{Arg}\left(z\right)&=\text{undefined},x=0,y=0\end{align*})
Now let's consider the first question, starting with \(x>1\) and \(y\in\mathbb{R}\) (i.e. \(\text{Re}\left(z-1\right)=x-1>0\) and \(\text{Re}\left(z-i\right)=x>0\)):
&=\text{Arg}\left(\left(x-1\right)+yi\right)\\&=\tan^{-1}{\left(\frac{y}{x-1}\right)}\\\text{Arg}\left(z-i\right)&=\text{Arg}\left(x+\left(y-1\right)i\right)\\&=\tan^{-1}{\left(\frac{y-1}{x}\right)}\\\text{Arg}\left(z-1\right)&=\text{Arg}\left(z-i\right)\\\therefore\tan^{-1}{\left(\frac{y}{x-1}\right)}&=\tan^{-1}{\left(\frac{y-1}{x}\right)}\\\tan{\left(\tan^{-1}{\left(\frac{y}{x-1}\right)}\right)}&=\tan{\left(\tan^{-1}{\left(\frac{y-1}{x}\right)}\right)}\\\frac{y}{x-1}&=\frac{y-1}{x},x\ne\lbrace0,1\rbrace\\xy&=\left(x-1\right)\left(y-1\right)\\xy&=xy-y-x+1\\0&=-x-y+1\\y&=1-x,x>1,y\in\mathbb{R}\end{align*})
Now let's consider \(0<x<1\) and \(y\in\mathbb{R}\) (i.e. \(\text{Re}\left(z-1\right)=x-1<0\) and \(\text{Re}\left(z-i\right)=x>0\)):
&=\text{Arg}\left(\left(x-1\right)+yi\right)\\&=\pi+\tan^{-1}{\left(\frac{y}{x-1}\right)}\\\text{Arg}\left(z-i\right)&=\text{Arg}\left(x+\left(y-1\right)i\right)\\&=\tan^{-1}{\left(\frac{y-1}{x}\right)}\\\text{Arg}\left(z-1\right)&=\text{Arg}\left(z-i\right)\\\therefore\pi+\tan^{-1}{\left(\frac{y}{x-1}\right)}&=\tan^{-1}{\left(\frac{y-1}{x}\right)}\\\tan^{-1}{\left(\frac{y-1}{x}\right)}&\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)\\\pi+\tan^{-1}{\left(\frac{y}{x-1}\right)}&\in\left(-\frac{\pi}{2}=\pi,\frac{\pi}{2}+\pi\right)\\&=\left(\frac{\pi}{2},\pi\right)\\\therefore\pi+\tan^{-1}{\left(\frac{y}{x-1}\right)}&=\tan^{-1}{\left(\frac{y-1}{x}\right)}\text{ has no solution}\end{align*})
Now let's consider \(x<0\) and \(y\in\mathbb{R}\) (i.e. \(\text{Re}\left(z-1\right)=x-1<0\) and \(\text{Re}\left(z-i\right)=x<0\)):
&=\text{Arg}\left(\left(x-1\right)+yi\right)\\&=\pi+\tan^{-1}{\left(\frac{y}{x-1}\right)}\\\text{Arg}\left(z-i\right)&=\text{Arg}\left(x+\left(y-1\right)i\right)\\&=\pi+\tan^{-1}{\left(\frac{y-1}{x}\right)}\\\text{Arg}\left(z-1\right)&=\text{Arg}\left(z-i\right)\\\therefore\pi+\tan^{-1}{\left(\frac{y}{x-1}\right)}&=\pi+\tan^{-1}{\left(\frac{y-1}{x}\right)}\\\therefore\tan^{-1}{\left(\frac{y}{x-1}\right)}&=\tan^{-1}{\left(\frac{y-1}{x}\right)}\end{align*})
This has the same solution as the first case, where \(x>1\), but with a different domain.
That is, \(y=1-x,x<0\).
Finally, let's consider the singularities, firstly, where \(x=0\):
i\\\text{Arg}\left(z-1\right)&=\pi+\tan^{-1}{\left(-y\right)}\\\text{Arg}\left(z-i\right)&=\frac{\pi}{2},y>1\\\text{Arg}\left(z-i\right)&=-\frac{\pi}{2},y<1\\\text{Arg}\left(z-i\right)&=\text{undefined},y=1\\\pi+\tan^{-1}{\left(-y\right)}&\in\left(\frac{\pi}{2},\pi\right)\\\therefore\pi+\tan^{-1}{\left(-y\right)}&\ne\pm\frac{\pi}{2}\end{align*})
There is no solution for \(x=0\)
Lastly, let's consider \(x=1\):
i\\\text{Arg}\left(z-i\right)&=\tan^{-1}{\left(y-1\right)}\\\text{Arg}\left(z-1\right)&=\frac{\pi}{2},y>0\\\text{Arg}\left(z-1\right)&=-\frac{\pi}{2},y<0\\\text{Arg}\left(z-1\right)&=\text{undefined},y=0\\\tan^{-1}{\left(y-1\right)}&\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)\\\therefore\tan^{-1}{\left(y-1\right)}&\ne\pm\frac{\pi}{2}\end{align*})
There is no solution for \(x=1\).
Putting it all together, \(\text{Arg}\left(z-1\right)=\text{Arg}\left(z-i\right)\) is described by the line \(y=1-x\) with restricted domain \(x<0\cup x>1\) in the Argand plane, where \(\text{Re}\left(z\right)=x\) and \(\text{Im}\left(z\right)=y\).
The algebra in this problem is relatively straightforward, because \(\tan{\left(\pi+\theta\right)}=\tan{\left(-\pi+\theta\right)}=\tan{\left(\theta\right)}\) and you should be able to apply it to part b) (which I haven't looked at yet, but I'll try and come back to at some stage).
The biggest difficulty in these problems is identifying the domain. I think it would be much easier to try and identify appropriate regions in the Argand plane by drawing some test diagrams.
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