Find $z$ for $\arg(z- 1) = \pi/4$. A point $P$ representing the complex number $z$ moves in the Argrand diagram so it lies always in the region defined by $|z- 1| \le |z - i|$ and $|z- 2 - 2i| \le 1$. If $P$ describes the boundary of the this region then find the values of $z$ in the form $x + iy$ when $\arg(z - 1) = \pi/4$.
$(z - 1) = |z- 1|/\sqrt{2} * (1 + i) \implies z = |z - 1|/\sqrt{2} * (1 + i) + 1$ 
Now I wish to find $|z- 1|$. 
$|z - 1| = |z - i| \implies \Re(z) = \Im(z)$
and $(z - 2 - 2i)(z^* - 2 + 2i) = 1 \implies |z|^2 - 4 = 0 \implies |z| = 2$.
I am stuck. Any hints ? 
 A: I would use a different approach and think of the regions geometrically in the Argand plane. Here are some hints--I'll leave details to you.
The requirement $|z-1|\le |z-i|$ says that point $z$ is closer to point $1$ than to point $i$. The boundary is where those distances are equal, the perpendicular bisector of the line segment between $1$ and $i$. This is the line $y=x$.
The boundary of the second condition $|z-2-2i|\le 1$ is the circle with center $2+2i$ and radius $1$.
The locus of $\mathrm{arg}(z-1)=\pi/4$ is the ray starting from the point $1$ with the angle of inclination $\pi/4$, i.e. $45°$. This is the part of the line $x=y+1$ above the real line ($x$-axis).
The first and third requirements cannot be satisfied simultaneously. The second and third can: solve the simultaneous equations for the circle and ray:
$$(x-2)^2+(y-2)^2=1$$
$$x=y+1$$
Substitute the second equation into the first and solve the quadratic equation in $y$. You get two solutions, and for each use the second equation to get the corresponding values of $x$. That gives you two points (complex value solutions).
You can easily verify that the equalities for requirements 2 and 3 are satisfied as well as the inequality in requirement 1.
By the way, just sketching the graphs of the three requirements is easy and finds the two solutions very quickly.
A: Let $z=x+yi$. 
Hence, $z-1=x-1+yi$.
Since $1=\tan{\arg(z-1)}=\frac{y}{x-1}$, we obtain $y=x-1$ for $x\neq1$.
