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Complex number problem

Pranav

Well-known member
Nov 4, 2013
428
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? :confused:

Any help is appreciated. Thanks!
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
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
Feb 15, 2012
1,967
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
Nov 4, 2013
428
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
Feb 15, 2012
1,967
If $z \in \Bbb R$ what possible values can $\arg(z)$ have?
 

Pranav

Well-known member
Nov 4, 2013
428
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. :confused:
 

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
Not ANY multiple, an INTEGER multiple.

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

Pranav

Well-known member
Nov 4, 2013
428
Not ANY multiple, an INTEGER multiple.

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

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

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
Does the magnitude of a complex number affect its angle?
 

Pranav

Well-known member
Nov 4, 2013
428
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

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
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
Nov 4, 2013
428
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? :confused:

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
Jan 17, 2013
1,667
$$arg(z(1+\bar{z}))-arg(z-|z|^2)=0$$

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
Nov 4, 2013
428
$$arg(z(1+\bar{z}))-arg(z-|z|^2)=0$$

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. :confused:
 

ZaidAlyafey

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MHB Math Helper
Jan 17, 2013
1,667

Pranav

Well-known member
Nov 4, 2013
428
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. :confused:
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
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
Nov 4, 2013
428
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? :confused:
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
Umm...but how does that answer my question in post #12? :confused:
I must be interpreting your question wrongly. It would be great if you rephrase it a little bit .
 

Pranav

Well-known member
Nov 4, 2013
428
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? :confused:
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
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? :confused:
You only proved that $z$ is real .
 

Pranav

Well-known member
Nov 4, 2013
428
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
Jan 17, 2013
1,667
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
Nov 4, 2013
428
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! :)