Simplifying the Argument of a Complex Number

[songoku]

Homework Statement

Please see below

Relevant Equations

arg (ab) = arg (a) + arg (b)

arg(a/b) = arg (a) - arg (b)

Let z = x + iy

$$\arg \left(\frac{1+z^2}{1 + \bar z^{2}}\right)=\arg (1+z^2) - \arg (1 + \bar z^{2})$$
$$=\arg (1+x^2+i2xy-y^2)-\arg(1+x^2-i2xy+y^2)$$

Then I stuck.

I also tried:
$$\frac{1+z^2}{1 + \bar z^{2}}=\frac{1+x^2+i2xy-y^2}{1+x^2-i2xy+y^2}$$

But also stuck

How to do this question? Thanks

 

[WWGD]

Your first identity needs to adjust if the sum of arguments is larger than $$2\pi$$.

 

[Delta2]

I think one helpful identity would be that $$\arg \bar z+\arg z=\arg(z\bar z)=\arg(|z|^2)=0 \pmod {2\pi}\Rightarrow \arg \bar z=2\pi-\arg z$$ and also $$\arg (z^n)=n\arg z \pmod {2\pi}$$.

But I can't figure out how we going to relate $\arg (1+z^2)$ to $\arg(z^2)$ I think here we must use the given fact that $|z|=1$ somehow...

 

[WWGD]

I think one helpful identity would be that $$\arg \bar z+\arg z=\arg(z\bar z)=\arg(|z|^2)=0\Rightarrow \arg \bar z=-\arg z$$ and also $$\arg (z^n)=n\arg z$$.

But I can't figure out how we going to relate $\arg (1+z^2)$ to $\arg(z^2)$ I think here we must use the given fact that $|z|=1$ somehow...

Remember that standard arguments, at least in my experience, have a " scope" of $$2\pi$$, so that, if, e.g., arg(z)=$$\pi$$, then arg$z^4 \neq 4\pi$

 

[Delta2]

yes right I think the arithmetic with arguments is modulo $2\pi$ right?

 

[Delta2]

Another idea since $|z|=1$ to set $z=e^{i\phi}$, $\bar z=e^{-i\phi}$ $\frac{\pi}{6}\leq\phi\leq\frac{\pi}{3}$ and see how far you can get with that fraction...

 

[pbuk]

Makes me think about the double angle formulae.

 

[Delta2]

I solved this by setting $z=e^{i\phi}=\cos\phi+i\sin\phi$ and working through complex algebra and trigonometry, but the possible answers match more than one. Does the question allows for this?

 

[Hall]

If the argument is between 30 and 60 degrees, and the length is one, can we assume the given complex number is oriented at 45 degress and real and complex coefficients are same? (What’s their length?)

Once you get z in explicit form by the above method you will have the answer, and then you can manipulate your algebra to get that correct answer.

 

[pbuk]

the possible answers match more than one. Does the question allows for this?

It looks like this is the case, yes.

 

[WWGD]

Arguments are defined in $$[a, a+2\pi)$$ in order to be continuous, or even well-defined, as $$ e^{\theta}=e^{\theta + 2\pi} $$ for any Real a. It took me a while to understand that's what was meant by a 'Branch" : a local inverse of the Exponential $exp z$. Wish I had known it earlier, it would have saved me a lot of time!

 

[SammyS]

From Post #1:

Let z = x + iy

$$\arg \left(\frac{1+z^2}{1 + \bar z^{2}}\right)=\arg (1+z^2) - \arg (1 + \bar z^{2})$$
$$=\arg (1+x^2+i2xy-y^2)-\arg(1+x^2-i2xy+y^2)$$

Then I stuck.

I also tried:
$$\frac{1+z^2}{1 + \bar z^{2}}=\frac{1+x^2+i2xy-y^2}{1+x^2-i2xy+y^2}$$

But also stuck

How to do this question? Thanks

You have some errors here.
If $z=x+iy$, then $\bar z=x-iy$, but $(x-iy)^2=x^2 -2ixy-y^2$.

Also, $|z|=1 \text{, so } x^2+y^2=1\text{, i.e. } 1-y^2 =x^2$ .

Using this in $1+z^2$ gives $2x^2+i2xy$, right?

Factor out $2x$.

Similarly for the denominator.

 

[songoku]

Remember that standard arguments, at least in my experience, have a " scope" of $$2\pi$$, so that, if, e.g., arg(z)=$$\pi$$, then arg$z^4 \neq 4\pi$

I learn that the principle value of argument for complex number is $-\pi < \theta \leq \pi$

If the argument is between 30 and 60 degrees, and the length is one, can we assume the given complex number is oriented at 45 degress and real and complex coefficients are same? (What’s their length?)

I don't think we can assume that

I solved this by setting $z=e^{i\phi}=\cos\phi+i\sin\phi$ and working through complex algebra and trigonometry, but the possible answers match more than one. Does the question allows for this?

From Post #1:

You have some errors here.
If $z=x+iy$, then $\bar z=x-iy$, but $(x-iy)^2=x^2 -2ixy-y^2$.

Also, $|z|=1 \text{, so } x^2+y^2=1\text{, i.e. } 1-y^2 =x^2$ .

Using this in $1+z^2$ gives $2x^2+i2xy$, right?

Factor out $2x$.

Similarly for the denominator.

I understand. I don't know whether the question allows more than one answer but I got that option (a), (b) and (c) can be the possible arguments.

Thanks you very much for the help and explanation WWGD, Delta2, pbuk, Hall, SammyS

 

[lurflurf]

I would let t=arg(z) then z=|z|exp(i t)=exp(i t)
$$\mathrm{arg}\left( \frac{1+z^2}{1+\bar{z}^2 }\right)=\mathrm{arg}\left( \frac{1+\exp(2i t)}{1+\exp(-2i t) }\right)$$
$$\mathrm{arg}\left( \frac{1+z^2}{1+\bar{z}^2 }\right)=\mathrm{arg}\left(\exp(2i t)\right)$$

 

[Orodruin]

Since this is already solved …

We have $\bar z (1+z^2) = z + \bar z$ and therefore $1+z^2 = (z + \bar z)/\bar z$ (and similarly for $1+\bar z^2 = (z+\bar z)/z$. It immediately follows that
$$
\frac{1+z^2}{1+\bar z^2} = \frac {z}{\bar z} = z^2,
$$
since $z \bar z=1$. Consequently, we have
$$
\arg\left(\frac{1+z^2}{1+\bar z^2}\right) = 2\arg z.
$$
No parametrizations necessary.

 

[Orodruin]

There is also a quite intuitive geometrical interpretation here based on the inscribed angle theorem.

Consider the circle of radius one in the complex plane centered at 1. Clearly, $1+z^2$ lies on this circle as |z|=1. The central angle between $z^2$ and the real line is 2 arg(z) and, since the origin is also on the circle, the inscribed angle between the real line and $1+z^2$ is half of this, ie, arg(z).

Since the denominator is the complex conjugate of the numerator, it directly follows that the argument of the ratio is 2 arg(z).

 

[FactChecker]

Since this is already solved …

We have $\bar z (1+z^2) = z + \bar z$ and therefore $1+z^2 = (z + \bar z)/\bar z$ (and similarly for $1+\bar z^2 = (z+\bar z)/z$. It immediately follows that
$$
\frac{1+z^2}{1+\bar z^2} = \frac {z}{\bar z} = z^2,
$$
since $z \bar z=1$. Consequently, we have
$$
\arg\left(\frac{1+z^2}{1+\bar z^2}\right) = 2\arg z.
$$
No parametrizations necessary.

Aha! Very good!
Another way to see that, EDIT It may be a little easier to see in the exponential form since |z|=1:
$(1+e^{i2\theta})/(1+e^{-i2\theta}) = (e^{i\theta}(e^{-i\theta}+e^{i\theta}))/(e^{-i\theta}(e^{i\theta}+e^{-i\theta})) = e^{i\theta}/e^{-i\theta} = e^{i2\theta}$

 

[Orodruin]

Aha! Very good!
Another way to see that, since |z|=1, is:
$(1+e^{i2\theta})/(1+e^{-i2\theta}) = (e^{i\theta}(e^{-i\theta}+e^{i\theta}))/(e^{-i\theta}(e^{i\theta}+e^{-i\theta})) = e^{i\theta}/e^{-i\theta} = e^{2i\theta}$

I believe this is what #14 did, just not writing out the intermediate steps.

 

[Orodruin]

Oh, same idea, just somewhat simpler, using $z\bar z = 1$:
$$
1+z^2 = z\bar z(1+z^2) = z(\bar z + z^2 \bar z) = z(z+\bar z).
$$
Now, $z + \bar z$ is real so $\arg(1+z^2) = \arg(z)$ (as long as $z+\bar z \neq 0$).

 

[FactChecker]

I believe this is what #14 did, just not writing out the intermediate steps.

Yes, I just realized that fact. I think it is a little easier to see in the exponential form and will edit it to make it clear that it is just a restatement.