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    I have a feeling the 'Undergraduate' :ahhhhh: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! ^^
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    (Original post by PhysicsGal)
    I have a feeling the 'Undergraduate' :ahhhhh: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?
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    (Original post by notnek)
    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 :/

    Also, another qu if it's okay:

    What is (-1 + i) * [(1/sqrt2) + (1/sqrt2)i] = to?

    I converted the 2 complex numbers into polar form and got sqrt2. But when I tried multiplying them in cartesian form, I kept coming up with -2/sqrt2, (= 2*sqrt2), which is wrong :/

    Thank you very much
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    (Original post by PhysicsGal)
    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: \disp|a+bi|=\sqrt{a^2+b^2}. 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.

    Edit Using \displaystyle \left |\frac{z_1}{z_2}\right | = \frac{|z_1|}{|z_2|} is faster here, which you realised yourself.
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    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 ^^)
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    (Original post by notnek)
    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: \disp|a+bi|=\sqrt{a^2+b^2}. 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.

    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 :/
    (Original post by aznkid66)
    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 ^^)
    Well...I got the polar form of (-1 + i) as what you got. The thing is, I got that the answer (which is right) to the full multiplication is sqrt2 (after converting [(1/sqrt2) + (i/sqrt2)] into polar form too, where theta = pi/4 I think..and modulus is sqrt1 = 1)...

    Yet when I tried multiplying those 2 complex numbers in their cartesian forms (eg x+iy) I get 2sqrt2 as the answer, which is wrong.
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    (Original post by PhysicsGal)
    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 :/
    \displaystyle |x-iy| = |x+(-y)i| = \sqrt{x^2+(-y)^2} = \sqrt{x^2+y^2}
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    Okay, so for this equation (I'm sorry I ain't using latex...it just takes forevs to learn ):

    (-1 + i) is multiplied with [(1/sqrt2) + (i/sqrt2)]

    Using the FOIL method (First, Outside, Inside, Last)
    = (-1/sqrt2) + (-i/sqrt2) + (i/sqrt2) + (i^2 / sqrt2)

    The i/sqrt2 cancels with the -i/sqrt2, and i^2 = -1, to give:
    -1/sqrt2 - 1/sqrt2
    = -2/sqrt2

    If I multiply by sqrt2, I get 2sqrt2, but it is essentially the same thing :/

    How come I get a different answer when I multiply the same 2 complex numbers in cartesian form and then in polar form?

    Thanks!
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    (Original post by notnek)
    \displaystyle |x-iy| = |x+(-y)i| = \sqrt{x^2+(-y)^2} = \sqrt{x^2+y^2}
    :eek:As if I had to go make a mistake on an easy bit!:awesome:It's silly mistakes like that that are making this seem so hard!:cry2:

    Thank you!
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    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}} = ?
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    (Original post by aznkid66)
    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 :/
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    Oh wait, I get you!
    If I multiply -2/sqrt2 by sqrt2/sqrt2 I get (-2sqrt2)/2 and the 2s cancel out, to give -sqrt2!!

    Genious!!

    Thanks
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    (Original post by PhysicsGal)
    \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)
    Yup, so the answer is not positive sqrt(2) but...?


    \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 :/

    It didn't, I just took the negative out (it's on the left of the fraction now).

    Do you understand that 2=\sqrt{2} \times \sqrt{2} ?

    Also, to clarify, your initial mistake was that you multiplied your answer by sqrt(2) and said that the answer times sqrt(2) wasn't equal to the answer...which isn't that surprising, haha..

    EDIT: I'm glad you got it ^^
 
 
 
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