Complex numbers: if (u+v)/(u-v) is purely imaginary, show that mod(u)=mod(v)

In summary: If I do multiply by ##u - v## as a first step, I will get ##\frac {u^2+2uv+v^2}{u^2-v^2} = ai##Do I now expand ##u+v=a+ib## and ##u-v=c+id## in the above equation now?Yes. And here I substituted ##u=p+iq## and ##v=r+is##, which means ##a= p+r## and ##c=p-q## etc. since we had ##u+v=a+ib\, , \,u-v=c+id##. Now replace ##a,
  • #1
jisbon
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Homework Statement
If u and v are complex numbers such that u+v/u-v is purely imaginary, show that mod(u)=mod(v)
Relevant Equations
NIL
I'm kind of stuck over here at one part, but something is telling me that it might be wrong too :( Do assist, thanks.
1564799648981.png
 
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  • #2
I don't know what mod(u) is but my calculation got ##|u|=|v|##. I started with ##u+v=a+ib## and ##u-v=c+id## and only wrote ##u,v## as single numbers as I had the condition of pure imaginary derived.
 
  • #3
fresh_42 said:
I don't know what mod(u) is but my calculation got ##|u|=|v|##. I started with ##u+v=a+ib## and ##u-v=c+id## and only wrote ##u,v## as single numbers as I had the condition of pure imaginary derived.
oh my bad mod(u) is |u|,
and what do you mean by writing u and v as single numbers?

EDIT: I got this after following your advice:

1564802998804.png

what should I do next ? hm
 
  • #4
There is a sign error. You have to expand the quotient by ##1=\dfrac{c-id}{c-id}##. Next you separate the real part (with corrected sign), since this is all we are interested in. We only know that this part is zero. Now a quotient is zero if ... Now you can replace ##u=p+iq## and ##v=r+is##. The advantage is, that we got rid of all unnecessary parts: the imaginary part and the denominator.

You set the imaginary part zero, but the real part is!
 
  • #5
fresh_42 said:
There is a sign error. You have to expand the quotient by ##1=\dfrac{c-id}{c-id}##. Next you separate the real part (with corrected sign), since this is all we are interested in. We only know that this part is zero. Now a quotient is zero if ... Now you can replace ##u=p+iq## and ##v=r+is##. The advantage is, that we got rid of all unnecessary parts: the imaginary part and the denominator.

You set the imaginary part zero, but the real part is!
Oh sh**. Yep I made a careless mistake right here. Not sure what you meant by replacing ##u=p+iq## and ##v=r+is## tho :( I'm still stuck at:

##\dfrac {a+ib}{c+id}## x ##\dfrac {c-id}{c-id}## = ##\dfrac {ac+bd}{c^2+d^2}## + ##\dfrac {i(bc-ad)}{c^2+d^2}##

Since imaginary, no real numbers.

So ##\dfrac {ac+bd}{c^2+d^2} = 0##

ac+bd=0

What should I do next?
 
  • #6
An alternative is to start with:

##\frac{u+v}{u-v} = ai##

For some real number ##a##. That seems a much simpler approach.
 
Last edited:
  • #7
PeroK said:
An alternative is to start with:

##\frac{u+v}{u-v} = ai##

For some real number ##a##. That seems a much simpler approach.
So something like: ##\frac{u+v}{u-v} *\frac{u+v}{u+v}## ? What difference will it be though compared to the one I did above a.k.a
##\dfrac {a+ib}{c+id}## x ##\dfrac {c-id}{c-id}## = ##\dfrac {ac+bd}{c^2+d^2}## + ##\dfrac {i(bc-ad)}{c^2+d^2}##

Since imaginary, no real numbers.

So ##\dfrac {ac+bd}{c^2+d^2} = 0##

ac+bd=0
 
  • #8
jisbon said:
So something like: ##\frac{u+v}{u-v} *\frac{u+v}{u+v}## ? What difference will it be though compared to the one I did above a.k.a
##\dfrac {a+ib}{c+id}## x ##\dfrac {c-id}{c-id}## = ##\dfrac {ac+bd}{c^2+d^2}## + ##\dfrac {i(bc-ad)}{c^2+d^2}##

Since imaginary, no real numbers.

So ##\dfrac {ac+bd}{c^2+d^2} = 0##

ac+bd=0

Why not simply multiply by ##u - v## as a first step?

There's no need to expand ##u, v## into Cartesian form.
 
  • #9
PeroK said:
Why not simply multiply by ##u - v## as a first step?

There's no need to expand ##u, v## into Cartesian form.

If I do multiply by ##u - v## as a first step, I will get ##\frac {u^2+2uv+v^2}{u^2-v^2} = ai##
Do I now expand ##u+v=a+ib## and ##u-v=c+id## in the above equation now?
 
  • #10
jisbon said:
If I do multiply by ##u - v## as a first step, I will get ##\frac {u^2+2uv+v^2}{u^2-v^2} = ai##
Do I now expand ##u+v=a+ib## and ##u-v=c+id## in the above equation now?
That's not what I get. I get

## u + v = (u-v)ai##

Sorry, I'm not getting alerts on my phone. So I'm missing replies.
 
  • #11
jisbon said:
Oh sh**. Yep I made a careless mistake right here. Not sure what you meant by replacing ##u=p+iq## and ##v=r+is## tho :( I'm still stuck at:

##\dfrac {a+ib}{c+id}## x ##\dfrac {c-id}{c-id}## = ##\dfrac {ac+bd}{c^2+d^2}## + ##\dfrac {i(bc-ad)}{c^2+d^2}##

Since imaginary, no real numbers.

So ##\dfrac {ac+bd}{c^2+d^2} = 0##

ac+bd=0

What should I do next?
Yes. And here I substituted ##u=p+iq## and ##v=r+is##, which means ##a= p+r## and ##c=p-q## etc. since we had ##u+v=a+ib\, , \,u-v=c+id##. Now replace ##a,b,c,d## by ##p,q,r,s## accordingly and calculate ##ac+bd=0## with the new variables, sort the positive summands on different sides of the equation and done.

I would not change the strategy in the middle of the race. With ##ac+bd=0## your are almost done. The rest is easy and fast.
 
  • #12
fresh_42 said:
Yes. And here I substituted ##u=p+iq## and ##v=r+is##, which means ##a= p+r## and ##c=p-q## etc. since we had ##u+v=a+ib\, , \,u-v=c+id##. Now replace ##a,b,c,d## by ##p,q,r,s## accordingly and calculate ##ac+bd=0## with the new variables, sort the positive summands on different sides of the equation and done.

I would not change the strategy in the middle of the race. With ##ac+bd=0## your are almost done. The rest is easy and fast.
But, so many unnecessary variables!
 
  • #13
PeroK said:
That's not what I get. I get

## u + v = (u-v)ai##

Sorry, I'm not getting alerts on my phone. So I'm missing replies.
It's okay. Thanks for your help.
As to the question could you check this:
##\frac{u+v}{u-v} *\frac{u+v}{u+v}## = ##\frac{u^2+2uv+v^2}{u^2-v^2} =ai##
How did you get:
##u+v=(u−v)ai##

EDIT: Nevermind my bad, I'm seeing this now. So if ##u+v=(u−v)ai## , what should I do next?
 
  • #14
fresh_42 said:
Yes. And here I substituted ##u=p+iq## and ##v=r+is##, which means ##a= p+r## and ##c=p-q## etc. since we had ##u+v=a+ib\, , \,u-v=c+id##. Now replace ##a,b,c,d## by ##p,q,r,s## accordingly and calculate ##ac+bd=0## with the new variables, sort the positive summands on different sides of the equation and done.

I would not change the strategy in the middle of the race. With ##ac+bd=0## your are almost done. The rest is easy and fast.
With ##ac+bd=0##, what should I do next?
 
  • #15
jisbon said:
With ##ac+bd=0##, what should I do next?
##u+v=a+ib = (p+iq)+(r+is) \text{ and } u-v=c+id = (p+iq)-(r+is)## means ##a=p+r\, , \,b=q+s\, , \,c=p-r\, , \,d=q-s## and now simply calculate ##ac+bd=0=(p+r)(p-r)+(q+s)(q-s)## and rearrange the terms.
 
  • #16
jisbon said:
It's okay. Thanks for your help.
As to the question could you check this:
##\frac{u+v}{u-v} *\frac{u+v}{u+v}## = ##\frac{u^2+2uv+v^2}{u^2-v^2} =ai##
How did you get:
##u+v=(u−v)ai##

EDIT: Nevermind my bad, I'm seeing this now. So if ##u+v=(u−v)ai## , what should I do next?
I would rearrange that to get an equation relating ##u## and ##v##.
 
  • #17
fresh_42 said:
##u+v=a+ib = (p+iq)+(r+is) \text{ and } u-v=c+id = (p+iq)-(r+is)## means ##a=p+r\, , \,b=q+s\, , \,c=p-r\, , \,d=q-s## and now simply calculate ##ac+bd=0=(p+r)(p-r)+(q+s)(q-s)## and rearrange the terms.
Oh my, finally gotten this actually. Is there actually a simpler way or is the clearest way to do it?
 
  • #18
jisbon said:
Oh my, finally gotten this actually. Is there actually a simpler way or is the clearest way to do it?
Yes, @PeroK's. Rearrange ##u+v=(u-v)ai## with ##u's## on one side, ##v's## on the other and take the norm aka modulus aka length aka absolute value, whatever you call it. You need that it is multiplicative.
 
  • #19
fresh_42 said:
Yes, @PeroK's. Rearrange ##u+v=(u-v)ai## with ##u's## on one side, ##v's## on the other and take the norm aka modulus aka length aka absolute value, whatever you call it. You need that it is multiplicative.
Ah alright :) Thanks so much for the help everybody :)
 
  • #20
jisbon said:
Ah alright :) Thanks so much for the help everybody :)

Just to say that I think all the steps I took were the most obvious.

1) Write down what you have been given.

##\frac{u+v}{u-v} = ai##

2) Get rid of the fraction:

##u + v = (u-v) ai##

3) Rearrange, as you are looking for the equation ##|u|= |v|##:

##u(1-ai) = -v(1 + ai)##

4) Take the absolute value:

##|u||1-ai| = |v||1+ai|##

5) Note that ##|1-ai| = |1+ai|##

In each case, the next step looked the most obvious - the one to try first.
 
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  • #21
PeroK said:
Just to say that I think all the steps I took were the most obvious.

1) Write down what you have been given.

##\frac{u+v}{u-v} = ai##

2) Get rid of the fraction:

##u + v = (u-v) ai##

3) Rearrange, as you are looking for the equation ##|u|= |v|##:

##u(1-ai) = -v(1 + ai)##

4) Take the absolute value:

##|u||1-ai| = |v||1+ai|##

5) Note that ##|1-ai| = |1+ai|##

In each case, the next step looked the most obvious - the one to try first.
Wow this is a fast method haha. Thanks for your guidance!
 

FAQ: Complex numbers: if (u+v)/(u-v) is purely imaginary, show that mod(u)=mod(v)

1. What are complex numbers?

A complex number is a number that can be expressed in the form of a+bi, where a and b are real numbers and i is the imaginary unit, defined as the square root of -1.

2. What is the modulus of a complex number?

The modulus, or absolute value, of a complex number is its distance from the origin on the complex plane. It is calculated by taking the square root of the sum of the squares of the real and imaginary parts of the complex number.

3. How do you determine if a complex number is purely imaginary?

A complex number is purely imaginary if its real part is equal to 0. This means it can be written in the form of bi, where b is a non-zero real number.

4. How do you show that (u+v)/(u-v) is purely imaginary if mod(u)=mod(v)?

If mod(u)=mod(v), then the distance from the origin on the complex plane for both u and v is the same. This means that the numerator and denominator of (u+v)/(u-v) will have the same magnitude, but opposite signs, resulting in a purely imaginary number.

5. Can a complex number have a negative modulus?

No, the modulus of a complex number is always a positive real number. It represents the distance from the origin on the complex plane, which cannot be negative. If the modulus is negative, it is likely that there was an error in calculation.

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