Complex numbers: Prove that if ABS( (z1 - i*z2)/(z1 + i*z2) ) = 1, then...

[Quick]

The full questions is:

Prove that ifABS( (z1 - i*z2)/(z1 + i*z2) ) = 1
where z1 and z2 are potentially any numbers on the complex plane
THEN

z1/z2 is real

How do you factor with dividing these (binomials)?


The equation I am working on is:
|(z1-iz2)/(z1+iz2)|=1

I was told the next step is:
|(z1/z2-i)/(z1/z2+1)|=1


I am branching off a different thread where a similar equation came up but I was told to make a separate thread for it. It's really just a brain exercise that someone posted on another forum to see if people can figure it out. I can't really figure it out yet.

My initial thoughts are that the z1/z2's would cancel each other out but I am told this is not how it works.

IDK that much about complex numbers. So IDK how to view the problem in terms of z1 and z2 being complex numbers.

I had previously tried to solve for i, which put i = 0, but I am told this is wrong.

Help?

 

[Dr.Peterson]

How do you factor with dividing these (binomials)?

The equation I am working on is:
|(z1-iz2)/(z1+iz2)|=1

I was told the next step is:
|(z1/z2-i)/(z1/z2+1)|=1

I am branching off a different thread where a similar equation came up but I was told to make a separate thread for it. It's really just a brain exercise that someone posted on another forum to see if people can figure it out. I can't really figure it out yet.

My initial thoughts are that the z1/z2's would cancel each other out but I am told this is not how it works.

IDK that much about complex numbers. So IDK how to view the problem in terms of z1 and z2 being complex numbers.

I had previously tried to solve for i, which put i = 0, but I am told this is wrong.

I hope you realize that this problem is unrelated to the thread you tried to put it on. Any similar appearance is just superficial. And your question, "How do you factor with dividing these (binomials)?", makes no sense. There is really no factoring involved here.

I also hope you understand why you can't "cancel" the way you imagined; frankly, if you don't, then you are not ready for anything near this level of algebra. Canceling in fractions means removing a common factor of the entire numerator and the entire denominator, not just anything that appears somewhere in each.

But I'll give you a couple ideas, if you want to pursue it.

The start you were given is appropriate, but copied incorrectly; dividing numerator and denominator by |z2|, you get

|(z1/z2 - i)/(z1/z2 + i)|=1

To make the rest easier, you can define a new complex variable z = z1/z2, so you have to show that if |(z - i)/(z + i)| = 1, then z must be real.

This is actually not hard to show by thinking geometrically, if you imagine plotting z on the Argand (complex) plane. The equation says that z - i and z + 1 are the same distance from the origin. You can take it from there.

If you don't have that level of knowledge of complex numbers, or just want to do it algebraically, you might start by defining z as x + iy, where x and y are real, and then simplify the division the usual way (multiplying by the complex conjugate of the denominator). Your goal will be to show that y = 0.

Does either of those ideas get you started? If so, try it, and show what you are able to do.

 

[Quick]

I hope you realize that this problem is unrelated to the thread you tried to put it on. Any similar appearance is just superficial. And your question, "How do you factor with dividing these (binomials)?", makes no sense. There is really no factoring involved here.

I also hope you understand why you can't "cancel" the way you imagined; frankly, if you don't, then you are not ready for anything near this level of algebra. Canceling in fractions means removing a common factor of the entire numerator and the entire denominator, not just anything that appears somewhere in each.

This is what I was thinking in regards to the z1/z2 cancelling out:

We have | ( z1 ÷ z2 - i ) ÷ (z1 ÷ z2 + i ) | =1

From here I thought what I could do is switch the position of z1 and z2 in the second polynomial.

How I see it is:

\(\displaystyle |( z1 ÷ z2 - i ) [ 1 ÷ ( z1 ÷ z2 + i ) ] | = 1\) (laTex is failing me because it's like the second time I have tried to use it. I've been trying to get it to look like it's in fraction form for a while and can't figure it out.)

I can figure out what I was trying to say.. I think I was trying to say something about the reciprocal and how when you multiply across they cancel out. But that's wrong so I am not sure what I was thinking there.

But I'll give you a couple ideas, if you want to pursue it.

The start you were given is appropriate, but copied incorrectly; dividing numerator and denominator

by |z2|, you get

|(z1/z2 - i)/(z1/z2 + i)|=1

To make the rest easier, you can define a new complex variable z = z1/z2, so you have to show that if |(z - i)/(z + i)| = 1, then z must be real.

This is actually not hard to show by thinking geometrically, if you imagine plotting z on the Argand (complex) plane. The equation says that z - i and z + 1 are the same distance from the origin. You can take it from there.

If you don't have that level of knowledge of complex numbers, or just want to do it algebraically, you might start by defining z as x + iy, where x and y are real, and then simplify the division the usual way (multiplying by the complex conjugate of the denominator). Your goal will be to show that y = 0.

Does either of those ideas get you started? If so, try it, and show what you are able to do.

So basically, make a graph between real numbers and imaginary numbers?

I'm just going to give up at this point. I thought I would be able to figure this out because I thought it was going to be easy math, but I am just starting to look into complex numbers and stuff so I don't think I have the skill set to do this problem.

there seems to be a problem with the quote of the second and third quote blocks separating. For some reason when I try and make the second and third quote one quote, there is a end of quote and beginning of quote that inserts itself in the post and I have no control over this.

 

[Dr.Peterson]

This is what I was thinking in regards to the z1/z2 cancelling out:

We have | ( z1 ÷ z2 - i ) ÷ (z1 ÷ z2 + i ) | =1

From here I thought what I could do is switch the position of z1 and z2 in the second polynomial.

How I see it is:

\(\displaystyle |( z1 ÷ z2 - i ) [ 1 ÷ ( z1 ÷ z2 + i ) ] | = 1\) (laTex is failing me because it's like the second time I have tried to use it. I've been trying to get it to look like it's in fraction form for a while and can't figure it out.)

I can figure out what I was trying to say.. I think I was trying to say something about the reciprocal and how when you multiply across they cancel out. But that's wrong so I am not sure what I was thinking there.



So basically, make a graph between real numbers and imaginary numbers?

I'm just going to give up at this point. I thought I would be able to figure this out because I thought it was going to be easy math, but I am just starting to look into complex numbers and stuff so I don't think I have the skill set to do this problem.

there seems to be a problem with the quote of the second and third quote blocks separating. For some reason when I try and make the second and third quote one quote, there is a end of quote and beginning of quote that inserts itself in the post and I have no control over this.

It will be fun to come back to after you've done some graphing with complex numbers, and so on. It doesn't require really advanced knowledge, but you need a lot of confidence to tackle it!

I've seen vBulletin do the same sort of thing, and worse. You just have to let it happen, I guess.

As for LaTeX, I'm not great at it, but here's an attempt:

\(\displaystyle \left | \dfrac{\dfrac{z_1}{z_2}-i}{\dfrac{z_1}{z_2}+i} \right |=1\)

You can right-click on it to see what I entered between the [ tex ] and [\ tex ] (no spaces) markers.

I sometimes resort to a visual editing tool; one I've found is here.

 

[mmm4444bot]

Regarding the v-Bulletin snafu, you can try highlighting the quotes, starting with the paragraph above and ending with the paragraph below. Then cut and paste it back in, but (if you're using Windows) try CTRL-SHIFT-V, to paste as text. This sometimes works, to strip out hidden forum code that's gone goofy.