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

[jisbon]

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.

 

[fresh_42]

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.

 

[jisbon]

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:

what should I do next ? hm

 

[fresh_42]

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!

 

[jisbon]

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]

An alternative is to start with:

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

For some real number $a$. That seems a much simpler approach.

 

[jisbon]

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]

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.

 

[jisbon]

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]

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]

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]

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!

 

[jisbon]

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?

 

[jisbon]

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]

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]

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 $v$.

 

[jisbon]

$u+v=a+ib = (p+iq)+(r+is) \text{ and } u-v=c+id = (p+iq)-(r+is)$ 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]

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.

 

[jisbon]

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]

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.

 

[jisbon]

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!