Complex number problem

Pranav

Well-known member
Problem:
If $z$ is a complex number such that
$$\arg(z(1+\overline{z}))+\arg\left(\frac{|z|^2}{z-|z|^2}\right)=0$$
then

A)$\arg(\overline{z})=-\pi/2$

B)$\arg(z)=\pi/4$

C)$|\overline{z}|<1$

D)$\ln\left(\frac{1}{|z|}\right)\in (-\infty,\infty)$

Attempt:
From the fact that $|z|=z\overline{z}$, I simplified the given equation to the following:
$$\arg\left(\frac{1+\overline{z}}{1-\overline{z}}\right)=0$$
If the argument of a complex number is zero, then it is equal to its conjugate, hence
$$\frac{1+\overline{z}}{1-\overline{z}}=\frac{1+z}{1-z}$$
Solving gives me $z=\overline{z}$. What to do with this?

Any help is appreciated. Thanks!

ZaidAlyafey

Well-known member
MHB Math Helper
I have some concerns about the notations you are using. Are you defining $$\displaystyle \text{arg}(z)$$ as the principle argument? $$\displaystyle \text{arg}(z)$$ is a usually defined as multivalued function .

Deveno

Well-known member
MHB Math Scholar
Which complex numbers equal their conjugates?

This seems to be a "trick question" as 3 of the possible replies are totally irrelevant, and the correct answer doesn't even need any calculation.

Pranav

Well-known member
I have some concerns about the notations you are using. Are you defining $$\displaystyle \text{arg}(z)$$ as the principle argument? $$\displaystyle \text{arg}(z)$$ is a usually defined as multivalued function .
That isn't mentioned in the problem statement so I guess we have to go with the usual definition.

Which complex numbers equal their conjugates?
Real numbers are equal to their conjugates but that still doesn't give me the answer.

Deveno

Well-known member
MHB Math Scholar
If $z \in \Bbb R$ what possible values can $\arg(z)$ have?

Pranav

Well-known member
If $z \in \Bbb R$ what possible values can $\arg(z)$ have?
arg(z) can be any multiple of $\pi$ but I don't see how this helps.

Deveno

Well-known member
MHB Math Scholar
Not ANY multiple, an INTEGER multiple.

How does that square with the 4 choices the problem poses?

Pranav

Well-known member
Not ANY multiple, an INTEGER multiple.

How does that square with the 4 choices the problem poses?
Yes, integer multiple, sorry.

That rules out option A and B but how to check for the other two options?

Deveno

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MHB Math Scholar
Does the magnitude of a complex number affect its angle?

Pranav

Well-known member
Does the magnitude of a complex number affect its angle?
No. So that means $|z|\in (0,\infty)$? That gives me answer D but the given answer is C.

Deveno

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MHB Math Scholar
That *is* interesting. Obviously I have made an unwarranted leap of logic.

I believe we are OK up to this point:

$z = \overline{z}$.

So let's agree that $z = a \in \Bbb R$.

Our original problem then becomes:

$\arg(a(1+a)) = \arg\left(\dfrac{|a|^2}{a - |a|^2}\right)$

Now $a(1+a) = a^2 + a$.

If $|a| > 1$, this is positive, so its arg is 0, so the other arg must be 0.

However, $a - |a|^2 < a - a = 0$ for such $a$, which would make the second arg $\pi$.

I leave it to you to example the cases $a = \pm 1$.

Thus, it must be the case that $|a| = |z| = |\overline{z}| < 1$.

My apologies for not looking closer.

Pranav

Well-known member
That *is* interesting. Obviously I have made an unwarranted leap of logic.

I believe we are OK up to this point:

$z = \overline{z}$.

So let's agree that $z = a \in \Bbb R$.

Our original problem then becomes:

$\arg(a(1+a)) = \arg\left(\dfrac{|a|^2}{a - |a|^2}\right)$

Now $a(1+a) = a^2 + a$.

If $|a| > 1$, this is positive, so its arg is 0, so the other arg must be 0.

However, $a - |a|^2 < a - a = 0$ for such $a$, which would make the second arg $\pi$.

I leave it to you to example the cases $a = \pm 1$.

Thus, it must be the case that $|a| = |z| = |\overline{z}| < 1$.

My apologies for not looking closer.
Before proceeding with any of the above, why are you doing this?

We already came to the conclusion that $z$ is a real number so $|\overline{z}|$ can be anything. I don't understand why are we getting back to the original equation.

ZaidAlyafey

Well-known member
MHB Math Helper
$$arg(z(1+\bar{z}))-arg(z-|z|^2)=0$$

Since $z \neq 0$

$$arg\left( \frac{1+\bar{z}}{1-\bar{z}}\right)=0$$

$$\frac{1+\bar{z}}{1-\bar{z}}=a>0$$

$$1+\bar{z}=a-a \bar{z} \to \bar{z}=\frac{a-1}{a+1}$$

Since $$|\bar{z}|=|z| =\left |\frac{a-1}{a+1} \right| <1 \,\,\,; a\neq 1$$

Now if $z=0$ it is clear that $|z|<1$ so

$$|z|<1$$

Pranav

Well-known member
$$arg(z(1+\bar{z}))-arg(z-|z|^2)=0$$

Since $z \neq 0$

$$arg\left( \frac{1+\bar{z}}{1-\bar{z}}\right)=0$$

$$\frac{1+\bar{z}}{1-\bar{z}}=a>0$$

$$1+\bar{z}=a-a \bar{z} \to \bar{z}=\frac{a-1}{a+1}$$

Since $$|\bar{z}|=|z| =\left |\frac{a-1}{a+1} \right| <1 \,\,\,; a\neq 1$$

Now if $z=0$ it is clear that $|z|<1$ so

$$|z|<1$$
Thanks ZaidAlyafey!

But that still doesn't answer my question in my post #12.

ZaidAlyafey

Well-known member
MHB Math Helper
If the argument of a complex number is zero, then it is equal to its conjugate, hence ...
Can you prove that ?

Pranav

Well-known member
Can you prove that ?
Let $z=re^{i\theta}$, then $\overline{z}=re^{-i\theta}$. Since the argument is zero, $z=r$ and $\overline{z}=r$ but I don't see why you asked me this.

ZaidAlyafey

Well-known member
MHB Math Helper
Notice that the argument is zero then the complex number is actually a positive real number. So $$\displaystyle arg(z) =0$$ implies that $z =a>0$. But if you say that $z=\bar{z}$ that also works for negative real numbers.

Last edited:

Pranav

Well-known member
Notice that the argument is zero then the complex number is actually a positive real number. So $$\displaystyle arg(z) =0$$ implies that $z =a>0$. But if you say that $z=\bar{z}$ that also works for negative real numbers.
Umm...but how does that answer my question in post #12?

ZaidAlyafey

Well-known member
MHB Math Helper
Umm...but how does that answer my question in post #12?
I must be interpreting your question wrongly. It would be great if you rephrase it a little bit .

Pranav

Well-known member
I must be interpreting your question wrongly. It would be great if you rephrase it a little bit .
Ok.

I reached the result that $z=\overline{z}$ and hence, $z$ can be any real number. This mean $|z|$ can be anything. Through your solution, you showed that $|z|<1$ but why do you get a different result than mine and also, how do we know which result is correct? What's wrong with my result?

ZaidAlyafey

Well-known member
MHB Math Helper
Ok.

I reached the result that $z=\overline{z}$ and hence, $z$ can be any real number. This mean $|z|$ can be anything. Through your solution, you showed that $|z|<1$ but why do you get a different result than mine and also, how do we know which result is correct? What's wrong with my result?
You only proved that $z$ is real .

Pranav

Well-known member
You only proved that $z$ is real .
So, from $z=\overline{z}$, we cannot comment on $|z|$. For that we have to get back to the original equation, right?

ZaidAlyafey

Well-known member
MHB Math Helper
So, from $z=\overline{z}$, we cannot comment on $|z|$. For that we have to get back to the original equation, right?
Exactly, $$\displaystyle z=\bar{z}$$ tells us that $z$ is a purely real complex number but we have no idea about its modulus.

Pranav

Well-known member
Exactly, $$\displaystyle z=\bar{z}$$ tells us that $z$ is a purely real complex number but we have no idea about its modulus.
Thanks ZaidAlyafey and Deveno!