Help in justifying complex number question

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September 3rd 2012, 11:50 AM
tempq1

Help in justifying complex number question

Given Z₁and Z₂ are two distinct complex numbers.
If |Z₁| = |Z₂|, then Re $\frac{z1+ z2}{z1 - z2}$ = 0
Is the above statement true or false?
Show working to justify your answer.
i managed to worked it out by letting z1= a+jb and z2= x+jy and then sub it in to work out and got true, not sure if i'm correct. Help is much appreciated (Nod)
September 3rd 2012, 12:28 PM
Plato

Re: Help in justifying complex number question

Quote:

Originally Posted by tempq1

Given $Z_1$ and $Z_2$ are two distinct complex numbers.
If $|Z_1|=|Z_2|$ , then Re $\frac{z_1+ z_2}{z_1 - z_2}$ = 0
Is the above statement true or false?
Show working to justify your answer.
i managed to worked it out by letting z1= a+jb and z2= x+jy and then sub it in to work out and got true, not sure if i'm correct. Help is much appreciated

It is true. Messy to show.
September 3rd 2012, 01:16 PM
MaxJasper

Re: Help in justifying complex number question

Quote:

Originally Posted by tempq1

Given Z₁and Z₂ are two distinct complex numbers.
If |Z₁| = |Z₂|, then Re $\frac{z1+ z2}{z1 - z2}$ = 0
Is the above statement true or false?
Show working to justify your answer.
i managed to worked it out by letting z1= a+jb and z2= x+jy and then sub it in to work out and got true, not sure if i'm correct. Help is much appreciated (Nod)

Representing z1 & z2 in exponential form we have:

$\text{z1}\text{:=}r e^{i \theta }$
$\text{z2}\text{:=}r e^{i \omega }$

Then: $\Re\left(\frac{\text{z1}+\text{z2}}{\text{z1}-\text{z2}}\right) = \left(\sin ^2(\theta )+\cos ^2(\theta )-\sin ^2(\omega )-\cos ^2(\omega )\right)/\[-2 (-1+\text{Cos}(\theta -\omega ))]= 0$
September 3rd 2012, 02:06 PM
Plato

Re: Help in justifying complex number question

Quote:

Originally Posted by MaxJasper

Representing z1 & z2 in exponential form we have:
$\text{z_1}\text{:=}r e^{i \theta }$
$\text{z_2}\text{:=}r e^{i \omega }$
Then: $\Re\left(\frac{\text{z1}+\text{z2}}{\text{z1}-\text{z2}}\right) = \left(\sin ^2(\theta )+\cos ^2(\theta )-\sin ^2(\omega )-\cos ^2(\omega )\right)/\[-2 (-1+\text{Cos}(\theta -\omega ))]= 0$

You can note that:
$\frac{e^{i \theta }+ e^{i \omega}}{e^{i \theta }- e^{i \omega}}\cdot\frac{e^{-i\theta }- e^{-i\omega}}{e^{-i \theta }- e^{-i \omega}}=\frac{e^{i(\omega-\theta)} - e^{i(\theta- \omega)}}{|e^{i \theta }- e^{i \omega}|^2}$ and $\text{Re}\left(e^{i(\omega-\theta)} - e^{i(\theta- \omega)}\right)=0$
I said it is messy.
September 3rd 2012, 03:13 PM
Soroban

Re: Help in justifying complex number question

Hello, tempq1!

Quote:

$\text{Given: }\,z_1\text{ and }z_2\text{ are two distinct complex numbers.}$

$\text{If }|z_1| = |z_2|,\,\text{ then }Re\left(\frac{z_1+z_1}{z_1-z_2}\right) \:=\:0$

$\text{Is the above statement true or false?}$

$\text{Let: }\:\begin{Bmatrix}z_1 \:=\:a+bj \\ z_2 \:=\:c+dj \end{Bmatrix}$

$\text{We are told: }\:|z_1| \,=\,|z_2| \quad\Rightarrow\quad \sqrt{a^2+b^2} \,=\,\sqrt{c^2+d^2}$

. . . . $a^2+b^2 \:=\:c^2+d^2 \quad\Rightarrow\quad (a^2+b^2) - (c^2+d^2) \:=\:0$ .[1]

$\text{Consider: }\:\frac{z_1+z_2}{z_1-z_2} \;=\;\frac{(a+c) + (b+d)j}{(a-c) + (b-d)j}$

$\text{Rationalize: }\:\frac{(a+c) + (b+d)j}{(a-c)+(b-d)j}\cdot\frac{(a-c)-(b-d)j}{(a-c)-(b-d)j}$

. . $=\;\frac{(a+c)(a-c) - (a+c)(b-d)j + (a-c)(b+d)j + (b+d)(b-d)}{(a-c)^2 - (b-d)^2}$

. . $=\;\frac{(a^2-c^2) -(ab-ad+bc-cd)j + (ab + ad - bc - cd)j + (b^2-d^2)}{(a-c)^2 - (b-d)^2}$

. . $=\;\frac{(a^2-c^2 + b^2-d^2) + 2(ad - bd)j}{(a-c)^2 - (b-d)^2}$

. . $=\;\frac{(a^2+b^2)-(c^2+d^2)}{(a-c)^2 - (b-d)^2} + \frac{2(ad-bc)}{(a-c)^2 - (b-d)^2}j$

$\text{The real component is: }\:\frac{(a^2+b^2) - (c^2+d^2)}{(a-c)^2 - (b-d)^2}$

. . which, according to [1], equals 0.
September 3rd 2012, 03:39 PM
Plato

Re: Help in justifying complex number question

Quote:

Originally Posted by Soroban

Hello, tempq1!
$\text{Let: }\:\begin{Bmatrix}z_1 \:=\:a+bj \\ z_2 \:=\:c+dj \end{Bmatrix}$
$\text{We are told: }\:|z_1| \,=\,|z_2| \quad\Rightarrow\quad \sqrt{a^2+b^2} \,=\,\sqrt{c^2+d^2}$
. . . . $a^2+b^2 \:=\:c^2+d^2 \quad\Rightarrow\quad (a^2+b^2) - (c^2+d^2) \:=\:0$ .[1]
$\text{Consider: }\:\frac{z_1+z_2}{z_1-z_2} \;=\;\frac{(a+c) + (b+d)j}{(a-c) + (b-d)j}$
$\text{Rationalize: }\:\frac{(a+c) + (b+d)j}{(a-c)+(b-d)j}\cdot\frac{(a-c)-(b-d)j}{(a-c)-(b-d)j}$
. . $=\;\frac{(a+c)(a-c) - (a+c)(b-d)j + (a-c)(b+d)j + (b+d)(b-d)}{(a-c)^2 - (b-d)^2}$
. . $=\;\frac{(a^2-c^2) -(ab-ad+bc-cd)j + (ab + ad - bc - cd)j + (b^2-d^2)}{(a-c)^2 - (b-d)^2}$
. . $=\;\frac{(a^2-c^2 + b^2-d^2) + 2(ad - bd)j}{(a-c)^2 - (b-d)^2}$
. . $=\;\frac{(a^2+b^2)-(c^2+d^2)}{(a-c)^2 - (b-d)^2} + \frac{2(ad-bc)}{(a-c)^2 - (b-d)^2}j$
$\text{The real component is: }\:\frac{(a^2+b^2) - (c^2+d^2)}{(a-c)^2 - (b-d)^2}$
. . which, according to [1], equals 0.

Well I did say it is messy, but not that messy.
That is a notation nightmare.