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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 $z$.

V.G
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Munchies
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    It means that when you start from the origin and go to the "direction" of $-\frac{3 \pi}{4}$ (the angle that the line makes with the $x$-axis), you end up in some point. But then when you move $3$ to the right from that point, you end up in the point $z$. That's the geometric interpretation of the first relation. – Matti P. Aug 05 '20 at 12:45
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    In the complex plane, draw a line through $(3,0)$ with angle $\frac{-3\pi}{4}$. $z$ lies somewhere along this line. – acat3 Aug 05 '20 at 12:46
  • @MattiP. Thank you heaps. I appreciate the help. How would i be able to solve the question? I'm still a bit lost. – Munchies Aug 05 '20 at 12:54
  • @RezhaAdrianTanuharja Thank you as well. i appreciate the help. Could you also help me solve the question? – Munchies Aug 05 '20 at 12:55
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    $z$, $(-3,0)$, $(3,0)$ form a right isosceles triangle (look at graph from some answers below). Easy to see that $z=-3-6i$ – acat3 Aug 05 '20 at 13:12

5 Answers5

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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 $(\tan^{-1}-\pi)$.)

enter image description here

mjw
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  • Thank you for the answer. I kinda do see where it's going. So would the first step being turning the arguments into polar form? Sorry, i'm still learning so I might be a bit slow. But I appreciate the answer a lot! – Munchies Aug 05 '20 at 12:57
  • Polar form helps turn multiplications into additions, and exponents into multiplication. However, in this case, there are no multiplications or exponents, and the angles are “nice”, so we should just stick to Cartesian form. – Benjamin Wang Aug 05 '20 at 13:04
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    Okay @BenjaminWang, fair enough. But doesn't "argument" invoke some discussion of polar form. I mean, I don't want to get into an argument $\cdots$ – mjw Aug 05 '20 at 13:10
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    Given a point in the complex plane $z$, the polar form is $|z| e^{i \arg z}$. – mjw Aug 05 '20 at 14:12
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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:

enter image description here

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 ${-\frac{3\pi}{4}}$:

enter image description here

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:

enter image description here

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

enter image description here

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

enter image description here

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:

enter image description here

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):

enter image description here

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 ${z=x+iy=-3-6i}$.

Riemann'sPointyNose
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enter image description here

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.

Benjamin Wang
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  • Thank you. I appreciate this a lot <3 – Munchies Aug 05 '20 at 12:53
  • With this info, how would i be able to find out z? – Munchies Aug 05 '20 at 12:53
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    You’re welcome. You can either just look at my grid lines, or deduce the (Cartesian) equations of the two lines. For such “nice” angles, can you “see” the equations? In case that was not clear, write $z = x + i y $ and find two equations involving $x$ and $y$. – Benjamin Wang Aug 05 '20 at 12:56
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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.

enter image description here

Jamminermit
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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 $z= -3-6i$.

Bernard
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Quanto
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