Complex Numbers Arguments

Hi all,

I have a question about the following:
arg(z-1) = arg(z+1) + pi/4
Prove that this is part of the set of points for which |z-i|=sqrt(2).

I have drawn the circle of |z-i|=sqrt(2) but don't know how to get started with arg(z-1) = arg(z+1) + pi/4. Any tips would be really appreciated.
I have tried to draw a diagram but am unsure what to do next.

RQ

Reply 1

Prasiortle

13

Original post by RussianQuestion

Hi all,

I have a question about the following:
arg(z-1) = arg(z+1) + pi/4
Prove that this is part of the set of points for which |z-i|=sqrt(2).

I have drawn the circle of |z-i|=sqrt(2) but don't know how to get started with arg(z-1) = arg(z+1) + pi/4. Any tips would be really appreciated.
I have tried to draw a diagram but am unsure what to do next.

RQ

See https://math.stackexchange.com/questions/1609945/show-that-argz-1-argz1-pi-4

Reply 2

RussianQuestion

OP

14

Original post by Prasiortle

See https://math.stackexchange.com/questions/1609945/show-that-argz-1-argz1-pi-4

Hiya

Is the explanation there given clear to you?
If so, could you think of any other ways of going about the problem?

RQ

Reply 3

Prasiortle

13

Original post by RussianQuestion

Hiya

Is the explanation there given clear to you?
If so, could you think of any other ways of going about the problem?

RQ

Assuming you're referring to the second answer by Joey Zou (not the rather brief first answer by user301068), then yes, I would say it's very clear and detailed. As for other ways, you could write z=x+iy, and replace the args with arctans of expressions in x and y, but then you run into the problem that arg(x+iy) isn't simply arctans(y/x) - specifically, if x<0 and y>0, you have to then add pi, or if x<0 and y<0, you have to subtract pi. Thus in order to do the problem this way, you have to break into a few cases, and then you can apply the sum of arctans formula and finally do some messy algebra to find an algebraic equation of the locus. Personally, I think the geometric method explained in the solution I linked to is easier.

Original post by RussianQuestion

Hi all,

I have a question about the following:
arg(z-1) = arg(z+1) + pi/4
Prove that this is part of the set of points for which |z-i|=sqrt(2).

I have drawn the circle of |z-i|=sqrt(2) but don't know how to get started with arg(z-1) = arg(z+1) + pi/4. Any tips would be really appreciated.
I have tried to draw a diagram but am unsure what to do next.

RQ

Geometrically is the way to go. Otherwise,

You essentially have two half lines;

1. Goes through (1,0) with gradient

\tan(\arg(z-1))

2. Goes through (-1,0) with gradient

\tan(\arg(z+1))

We are looking for the general point where these two intersect. Let's denote it by

z=x+iy

. Then...

\tan[\arg(z-1)](x-1) = \tan[\arg(z+1)](x+1)

gives us the point of intersection along the real axis. Solving it...

\tan[\arg(z+1) + \frac{\pi}{4}](x-1) = \tan[\arg(z+1)](x+1)

\dfrac{\tan[\arg(z+1)]+\tan[\frac{\pi}{4}]}{1-\tan[\arg(z+1)]\tan[\frac{\pi}{4}]}(x-1) = \tan(\arctan[\frac{y}{x+1}])(x+1)

\dfrac{\frac{y}{x+1}+1}{1-\frac{y}{x+1}}(x-1) = y

Rearranging yields the eq. of a circle on the complex plane, you just need to rewrite it in terms of the

z

-modulus notation.

Reply 5

RussianQuestion

OP

14

How would I go about drawing a sketch of arg(z-1) = arg(z+1) + pi/4?

I apologise for the number of questions. This should be more intuitive, but I’m struggling to get used to representing complex numbers graphically.

Complex Numbers Arguments

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