Moduli and Conjugates of complex numbers ?

I have a feeling the 'Undergraduate' label is putting people off from responding.

Anyhoo:

If a question asks me to find the Real part of a complex number divided by it's conjugate, do I always get 1? The answer in the book says I get
[the square root of (x^2 - y^2)]/(x^2 + y^2)

O__O


Here, z is the complex conjugate of z (not sure how to get the bar on top of the z)

Example Qu: Find Re(z/z):

My answer: The real part of a complex number, z = x + iy, is x. The conjugate of z is x - iy, which also has a real part of x. x/x = 1. Thus the answer is always 1.

Another qu: Find, in terms of x & y, |(z/z)|:

Why does the answer for this = [the square root of (x^2 - y^2)]/(x^2 + y^2) (as found on the net)? Or does the answer = 1 (as the book says)?

Thanks! ^^

You're getting mixed up between $\displaystyle \Re \left(\frac{z}{\bar{z}}\right)$ and $\displaystyle \frac{\Re(z)}{\Re (\bar{z})}$. They are not the same. E.g.

$z=1+2i, \bar{z} = 1-2i$

$\displaystyle \frac{\Re(z)}{\Re(\bar{z})} = \frac{1}{1} =1$.

And this is true for all complex numbers as you showed.


But, $\displaystyle \frac{z}{\bar{z}} = \frac{1+i}{1-i} = \frac{(1+i)^2}{(1+i)(1-i)}=...=\frac{-3+4i}{5}$

So $\displaystyle \Re \left(\frac{z}{\bar{z}}\right) = \Re \left(\frac{-3+4i}{5}\right) = \frac{-3}{5}$


Can you see where you went wrong now?

Thank you, that does make sense

Do you know why the modulus of (a complex number/it's complex conjugate) is 1?

I worked it out by substituting some numbers for x and y, and it doesn't = 1..it changes depending on what x and y are :/

It does equal 1 so you must've made a mistake somewhere. Can you post working where you've shown that it isn't 1?

Are you able to express $\displaystyle \frac{z}{\bar{z}}$ in terms of x and y? If you managed to do the first question then you probably have.

(The answer you gave from the book is wrong for the first question. There should be no square root).

Once you've done that, you can use: . If you're still stuck, please post everything you've done and say where you're up to.

I can help you with your next question once you've done this one.

For the modulus and real part of a complex number divided by its conjugate, the algebra might be easier if you use polar form.

$\text{Let } z= x + i y = r(\cos \theta + i \sin \theta)=r\text{cis} \theta$

$\bar{z} =r(\cos \theta - i \sin \theta)=r(\cos (-\theta) + i \sin (-\theta)) = r\text{cis}(- \theta)$

$\dfrac{z}{\bar{z}} = \dfrac{ r\text{cis} \theta} {r\text{cis} (- \theta)} = \text{cis} 2 \theta = \cos 2 \theta + i \sin 2 \theta$

$x = r\cos \theta, y = r \sin \theta, x^2 + y^2=r^2$

$\left|\dfrac{z}{\bar{z}} \right| = |\text{cis} 2 \theta | = 1$

$\Re\left( \dfrac{z}{\bar{z}} \right) = \cos 2 \theta = \cos^2 \theta - \sin^2 \theta = \dfrac{ (r \cos \theta)^2 - (r \sin \theta)^2 } {r^2} = \dfrac{x^2 - y^2}{x^2 + y^2}$

---

As for the next problem... both your answers are wrong ._.

$|-1+i|=\sqrt{2}$

Thus:

$-1+i = \sqrt{2} \left( \dfrac{1}{\sqrt{2}} - \dfrac{1}{\sqrt{2}} i \right) = \sqrt{2} \text{ cis} \dfrac{3 \pi}{4}$

Did you get that? (Note: it'd probably be easier to identify your error if you show your working ^^)

Oh, I thought I had to divide the numbers first, and then do their modulus, but apparently the moduus OF the division of 2 numbers is the same as the modulus of 1 number divided by the modulus of another number.

Sooo, if I try it now:
|(x + iy)/(x - iy)|
= |(z+iy)| / |(x - iy)|
= [sqrt (x^2 + y^2)] / [sqrt(x^2 - y^2)]
And I think I'm stuck..
But if I substitute some numbers for x and y (as they are both real and not imaginary numbers)..such as x= 4, y = 3, I get:
sqrt25 / sqrt(7)
= 5 / sqrt7
I'm confused :/

You did it wrong in polar form ^^

$\dfrac{3 \pi}{4}+\dfrac{\pi}{4}= ?$

$\cos \left( \dfrac{3 \pi}{4}+\dfrac{\pi}{4} \right) + i \sin \left( \dfrac{3 \pi}{4}+\dfrac{\pi}{4} \right) = ?$

Also, you did it wrong in cartesian form too:

$\dfrac{-2}{\sqrt{2}} = -\dfrac{\sqrt{2} \times \sqrt{2} }{\sqrt{2}} = ?$

$\dfrac{3 \pi}{4}+\dfrac{\pi}{4}=$ = pi

$\cos \left( \dfrac{3 \pi}{4}+\dfrac{\pi}{4} \right) + i \sin \left( \dfrac{3 \pi}{4}+\dfrac{\pi}{4} \right) =$
= -1 + 0i
= -1
(sin of pi = 0 and cos of pi = -1)

$\dfrac{-2}{\sqrt{2}} = -\dfrac{\sqrt{2} \times \sqrt{2} }{\sqrt{2}} = ?$

Can I ask, where did you get -2/sqrt2 from?

Thanks

Edit: Derp me, I know where you got -2/sqrt2 from, sorry :P I don't get why the -2 disappeared from the top though..or what is happening with the last part of that equation :/