This article assumes basic knowledge of complex numbers, argand diagrams, subtraction of vectors.

The argument of a complex number `arg(z)`

is basically the angle which the vector makes with the horizontal, with the range `-π < arg(z) ≤ π`

.

A statement like `arg(z) = π/2`

would basically just mean a halfline starting from the origin, going vertically upwards.

So, then we came across the question `arg(z) = arg(z - 1 + i)`

and were asked to plot the answer. Eventually, we got the answer, but not everyone understood why it was the case. Even after an hours worth of discussion, there was still uncertainty. It did take me some time to grasp it as well.

I therefore decided to make this (hopefully, simple) guide to the problem.

———————————————

1)Rewrite as `arg(z) = arg(z - [-1 + i])`

2)As a subtraction of two vectors, we can find the relevant vectors on a graph as follows :

3)Now we have to find points where their arguments (the angles they make with the horizontal) are equal. Let us do this by picking some random point `P(x,y)`

and checking whether their arguments are equal.

4)As you can see, the arguments are not the same. So we shall try somewhere else.

5)Finally, we try one which yields some success.

6)As you can see, the two have the same argument. Also, the vectors overlap each other. We have found one point, and so we shall try another point. We shall leave all points of success on the graph, just for future reference.

7)Another success. The point P has just been moved a bit. As you might begin to notice, the points where the arguments are same lie on a straight line. In fact, with a few more examples, the points would form this :

8)In the figure above, we have also started trying on the other side (in the 4th quadrant). And we have a success there. In fact, some of the points where the arguments are same are as follows:

9)Joining up all the points into a line gives us the answer for this question :

A half-line from the origin and another half-line from the point (1, -1). There is no no line in between the two (refer to the figure before step 5 for explanation – They do not have the same argument there – At any point between the two points, the arrows will always be pointing in the opposite direction). Also, the points (0,0) and (1, -1) are not included (as there is no argument here)

Hmm…graphical trial-and-error is not very rigorous. Perhaps you should try an algebraical proof?

You could start with drawing the two vectors and the resultant on the Argand diagram, stating that the angle between the two complex numbers is zero and finding the conditions for it.

The whole point of this was to make it easier for people to ‘grasp’ the idea… not so much as to actually see why it happens. Algebraical proof is still more ‘precise’, or not so random, of course…

And for that kind of proof, how are you going to come to the conclusion except that arg(z) = arg(z-1)? There isn’t a ‘easier to understand’, say, cartesian form… Unless you have it in polar form? I don’t know… but how would that prove it?

Oh yeah… when you say draw the two vectors, resultant, and find the conditions for the arg to be same…. just wondering, is that not what I did?

I meant plot z and (-1 + i), and then your resultant will be (z – 1 + i). I think it’s easier to see the answer that way.

An algebraical proof:

We prove that arg(z) = arg(z – 1 + i).

Let z = a+bi.

For arg(z) = arg(z-1+i)

=> arg(a+bi) = arg(a-1 + (b+1)i)

These two conditions must hold:

(a-1)/a = (b+1)/b [1]

a is the same sign as a-1 [2]

From [1]

1 – 1/a = 1 + 1/b

-1/a = 1/b

a = -b

From [2]

a(a-1)>0

a1

Therefore your set of solutions for z is

{a-ai in C: a1}. Which is what you drew.

Oops…looks like the less-thans and greater-thans are gobbled up. The second condition is to make sure that both complex numbers are in the same quadrant (if not their arguments would differ by pi).

From [2]

a(a-1) GT 0

0 LT a LT 1

Set of solutions for z is {a-ai in C: 0 LT a LT 1}.

aiyah maths test over alreayd… relax lah…

Aargh…sorry…my correction is wrong

a LT 0 -or- a GT 1

Set is {a-ai in C: a LT 0 or a GT 1}