# 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)
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. Last edited by a moderator:
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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.

#### 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? 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!

#### 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

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

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

Last edited:

#### jisbon

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

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.

#### 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

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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!

#### 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

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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 $v$.

#### 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

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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.

#### 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

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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.

• Delta2 and ehild

#### 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!