What does $\arg(z-3)= \frac{-3\pi}{4}$ mean? How to visualize it? What does $\arg(z-3)= \frac{-3\pi}{4}$ mean? How to visualize it?
I can't seem to visualize what it looks like. Could someone please teach me this.
I'm currently trying to solve this question and want to break it down into pieces to understand the whole question:
Question: If $\arg(z-3)= \frac{-3\pi}{4}$ and $\arg(z+3)=\frac{-\pi}{2}$, find $z$.
 A: People have given you a nice graphical approach above - I'll try to give a slightly more in depth explanation of this approach, as well as give you another approach.
So, the argument of a complex number is essentially just the angle the number makes with regards to the positive real axis:

notice a negative angle means we go round clockwise, whereas a positive angle means we go round anti-clockwise.
Now, ${z-3}$ essentially means "take the complex number $z$, and then go left three units". When this is done, we are told the angle it makes with the positive real axis is ${-\frac{3\pi}{4}}$:

We don't know the magnitude of ${z-3}$, but all complex numbers with an angle of ${-\frac{3\pi}{4}}$ must lie somewhere on that line. So ${z-3}$ lies on that line. We don't know where, but it does somewhere. By moving the line across to the right ${3}$, we will have a line on which ${z}$ must lie:

We can apply exactly the same logic with the information ${\arg(z+3)=\frac{-\pi}{2}}$:

Moving this line left three, we will get a line again on which $z$ must lie:

And so now we have two lines which we know $z$ must lie on both. The intersection of these lines should then tell us what the value of $z$ is! This is because $z$ needs to satisfy being on both lines:

So now simply find the equation that represents both of the lines in question, and use some nice algebra to find out where they intersect. If you are struggling with this stage, upon request I can also explain how to do this.

Another quick (and a naughty) way to arrive at the answer is to set up a system of Algebraic equations. If ${z=x+iy}$, then you have
$${\arg((x-3) + iy) = \frac{-3\pi}{4}}$$
So what? Well, if ${z=x+iy}$, then ${\frac{y}{x}}$ will give you the tangent of the acute angle that the complex number makes with the pos/neg real axis (whether it's positive or negative depends on the quadrant the number lies in). Since ${-\frac{3\pi}{4}}$ is half way between ${-\frac{\pi}{2}}$ and ${-\pi}$, the acute angle it will make with the axis is ${\frac{\pi}{4}}$ (you can refer to the diagrams above once again to see what I mean):

So
$${\frac{y}{x-3}=\tan\left(\frac{\pi}{4}\right)=1}$$
Now, using
$${\arg((x+3)+iy)=-\frac{\pi}{2}}$$
will lead to a slight problem, since you will end up saying something like
$${\frac{y}{x+3}\stackrel{?}{=}\tan\left(\frac{\pi}{2}\right)}$$
(but - ${\tan(x)}$ "blows up to infinity" at ${x=\frac{\pi}{2}}$).
What's the solution to this? Well a bit of thought will lead you to ${x=-3}$. Now, subbing in ${x=-3}$ into the first equation gives
$${\frac{y}{-6}=1}$$
Which gives us ${y=-6}$. Hence ${z=x+iy=-3-6i}$.
A: 
Can you see what’s going on here? You can look up how to draw diagrams of points and lines in the complex plane.
To break it into pieces, think about what $\arg(u) = \theta$ means.
A: so the argument is simply the angle the complex number makes with the real axis when plotted on an argand diagram (note that a negative angle means you go clockwise from the positive real axis). In general arg(Z-a) is a half line starting at the point a. I have done a sketch to hopefully show this.

A: If $z=r e^{i\theta}$, then $\text{Arg}(z)= \theta + 2k\pi$, where $-\pi<\text{Arg}(z)\le \pi$.
So, $$\begin{aligned} 
\text{Arg}(z-3)= -\frac{3\pi}{4} &\Longrightarrow z-3 = r_1 e^\frac{-{i 3 \pi}}{4} \Longrightarrow z= 3+ r_1 e^\frac{-{i 3 \pi}}{4} \\
\text{Arg}(z+3)= -\frac{\pi}{2} &\Longrightarrow z+3 = r_2 e^\frac{-{i  \pi}}{2}\Longrightarrow z = -3 + r_2 e^\frac{-{i  \pi}}{2}\\
\end{aligned}$$
The first equation implies that $z$ lies on a ray starting from the point $3$ and directed at an angle $-\frac{3\pi}{4}$ (shown in yellow).  The second equation implies that $z$ lies on a ray starting from the point $-3$ and directed at an angle $-\frac{\pi}{2}$ (shown in blue).  The intersection is the point $z=-3-6i = \sqrt{45} e^{i \tan^{-1} 2 - i\pi}.$ (The green ray starts at the origin and ends at our point $z=-3-6i$.  It has magnitude $\sqrt{45}$ and argument $(\tan^{-1}-\pi)$.)

A: Note that $\operatorname{Arg}(z-3)=-\frac{3\pi}4$ is
$$x+i y-3=re^{\frac{-3\pi}{4}i }= -\frac r{\sqrt2}  -\frac r{\sqrt2} i 
$$
which leads to $x-3=  -\frac r{\sqrt2},\>\>\>y = -\frac r{\sqrt2} $, or
$$y=x-3$$
Thus, $\operatorname{Arg}(z-3)=-\frac{3\pi}4$ represents a line. Similarly, $\operatorname{Arg}(z+3)=-\frac{\pi}2$ represents $x=-3$. The two lines intersect at $z= -3-6i$.
