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

  • Thread starter jisbon
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301
25
Homework Statement
If u and v are complex numbers such that u+v/u-v is purely imaginary, show that mod(u)=mod(v)
Homework 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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fresh_42

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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.
 
301
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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
 

fresh_42

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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!
 
301
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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?
 

PeroK

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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:
301
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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
 

PeroK

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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.
 
301
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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?
 

PeroK

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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.
 

fresh_42

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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.
 

PeroK

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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!
 
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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?
 
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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?
 

fresh_42

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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.
 

PeroK

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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##.
 
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##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?
 

fresh_42

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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.
 
301
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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 :)
 

PeroK

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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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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!
 

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