Show that $\arg(z-1)=\arg(z+1) +\pi/4$. Prove that the set of points for which $\arg(z-1)=\arg(z+1) +\pi/4$ is part of the set of points for which $|z-j|=\sqrt 2$ and show which part clearly in a diagram.
Intuition for the geometrical interpretation of this would be really appreciated. 
And I was wondering greatest and lowest value for $|z+k|$ and why does this represent a circle ?
The question might be too basic but I just needed a intuition behind this. Concept of vectors is quite hard to follow.
Thanks in advance .
 A: Hint:
$|z-j|=\sqrt 2$ is the equation of the circle of center $j$ and radius $\sqrt 2$ i.e. $z=j+\sqrt 2e^{i\theta}$ for $\theta\in [0,2\pi[$.
I let you find $\theta$ s.t. $\arg(z-1)=\arg(z+1)+\pi/4$
A: Since $\arg(z-1) = \arg(z+1)+\frac{\pi}{4}$, we note that $\text{Im}(z-1) = \text{Im}(z+1)>0$ (since you have to rotate counterclockwise from $z+1$ to get to $z-1$).
Now, consider the points $z-1$ and $z+1$ as points in the plane, and let $A = z-1$ and $B=z+1$. Let $C=0$ be the origin, and consider the circumcircle of triangle $ABC$ (call it $\omega$). Since $\angle ACB = \frac{\pi}{4}$, the arc subtended by $AB$ in $\omega$ must cover an angle of $\frac{\pi}{2}$. Thus, if we let $O$ be the center of $\omega$, then $\angle AOB = \frac{\pi}{2}$. Since $AO = BO$, and $AB = |(z-1)-(z+1)| = 2$, it follows that $AO = BO=\sqrt{2}$ since $AOB$ is a 45-45-90 triangle. This implies that $CO = \sqrt{2}$ as well.
We also know the center of the circle, $O$, must lie on the perpendicular bisector of $AB$ which is just a vertical line through $z$. Furthermore, through easy calculations, we see that the distance between $z$ (the midpoint of $AB$) and $O$ must be $1$. So the center of the circle must be either $z+j$ or $z-j$. The fact that $CO = \sqrt{2}$ implies
$$|z\pm j| = \sqrt{2} \hspace{0.3cm} \text{(depending on whether the center is }z+j\text{ or }z-j\text{)}.$$
So we're almost done--we just need to show that we want a $-$ instead of a $+$ in the equation above. I'll do this by dividing this into cases:


*

*Case 1: Both $A$ and $B$ are in the same quadrant. If both are in the first quadrant, then the perpendicular bisector of $CA$ slopes downwards and intersects the bisector of $AB$ somewhere below $AB$. If both are in the second quadrant, apply the same reasoning to the bisector of $CB$.

*Case 2: $A$ and $B$ are in different quadrants. If $O$ lies above line $AB$, then $A$, $C$, and $B$ are in counterclockwise order along the circle, and they lie in the same half-circle below the horizontal diameter of the circle. But that implies $\angle ACB>\frac{\pi}{2}$, which is false.


Thus, the center of the circle must be below line $AB$, so the center is $z-j$, and from above we have $|z-j| = \sqrt{2}$. I will let you figure out which points satisfying $|z-j| = \sqrt{2}$ also satisfy $\arg(z-1) = \arg(z+1)+\frac{\pi}{4}$.
