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    hey can anyone help me with this one question?
    ok well here goes

    gven that z=2-i and z^2 = 3-4i

    hence or otherwise find the roots, z1 and z2 of the equation

    (z+i)^2 = 3-4i

    display these roots on an argand diagram
    (a) deduce tht |z1-z2| = 2(root5)
    b) find the value of arg(z1+z2)

    help!its in the p4 book by heineman page 37 question 23! its a past exam question and i cant do it!! :eek:
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    (Original post by kriztinae)
    hey can anyone help me with this one question?
    ok well here goes

    gven that z=2-i and z^2 = 3-4i

    hence or otherwise find the roots, z1 and z2 of the equation

    (z+i)^2 = 3-4i

    display these roots on an argand diagram
    (a) deduce tht |z1-z2| = 2(root5)
    b) find the value of arg(z1+z2)

    help!its in the p4 book by heineman page 37 question 23! its a past exam question and i cant do it!! :eek:
    it seems wrong
    (z+i)^2 = (2-i+i)^2 = 4 =/= 3-4i
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    (Original post by keisiuho)
    it seems wrong
    (z+i)^2 = (2-i+i)^2 = 4 =/= 3-4i
    how would you do it though?

    i tried getting rid of the i's using various substituition then came to z = (2-+root(2))/2
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    (Original post by keisiuho)
    it seems wrong
    (z+i)^2 = (2-i+i)^2 = 4 =/= 3-4i
    thank you!! thats what ithought!
    ill leave it! thanx
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    I think it should be:
    gven that if z=2-i then z^2 = 3-4i
    and the z in the equation is not necessarily 2-i

    so (z+i)^2 = 3-4i
    (z+i)^2 = (2-i)^2
    z+i = 2-i OR z+i = i-2
    z = 2-2i OR -2
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    (Original post by keisiuho)
    I think it should be:
    gven that if z=2-i then z^2 = 3-4i
    and the z in the equation is not necessarily 2-i

    so (z+i)^2 = 3-4i
    (z+i)^2 = (2-i)^2
    z+i = 2-i OR z+i = i-2
    z = 2-2i OR -2
    hmm maybe, ill keep trying
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    (Original post by kriztinae)
    hey can anyone help me with this one question?
    ok well here goes

    gven that z=2-i and z^2 = 3-4i

    hence or otherwise find the roots, z1 and z2 of the equation

    (z+i)^2 = 3-4i

    display these roots on an argand diagram
    (a) deduce tht |z1-z2| = 2(root5)
    b) find the value of arg(z1+z2)

    help!its in the p4 book by heineman page 37 question 23! its a past exam question and i cant do it!! :eek:
    (z+i)^2 = 3-4i
    z² + 2zi + i² = 3 - 4i
    z² + 2zi + 4i - 4 = 0
    z² + z(2i + 4) - 4 = 0
    (z + (i+2))² - (2i + 4)² - 4 = 0
    (z + (i+2))² - (4i² + 16i + 16) - 4 = 0
    (z + (i+2))² + 4 - 16i - 20 = 0
    (z + (i+2))² = 16(1 + i)
    z + (i+2) = ±4√(1 + i)
    z = i+2 ±4√(1 + i)

    z1 = i+2+4√(1 + i)
    z2 = i+2-4√(1 + i)

    |z1-z2| = |(i+2+4√(1 + i)) - (i+2-4√(1 + i))|
    |z1-z2| = |(8√(1 + i)| =/= 2√5

    lol, I don't think I'm ready for these P4 questions.
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    (Original post by mik1a)
    (z+i)^2 = 3-4i
    z² + 2zi + i² = 3 - 4i
    z² + 2zi + 4i - 4 = 0
    z² + z(2i + 4) - 4 = 0
    (z + (i+2))² - (2i + 4)² - 4 = 0
    (z + (i+2))² - (4i² + 16i + 16) - 4 = 0
    (z + (i+2))² + 4 - 16i - 20 = 0
    (z + (i+2))² = 16(1 + i)
    z + (i+2) = ±4√(1 + i)
    z = i+2 ±4√(1 + i)

    z1 = i+2+4√(1 + i)
    z2 = i+2-4√(1 + i)

    |z1-z2| = |(i+2+4√(1 + i)) - (i+2-4√(1 + i))|
    |z1-z2| = |(8√(1 + i)| =/= 2√5

    lol, I don't think I'm ready for these P4 questions.
    erm thanx!
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    (z+i)^2 = 3 - 4i
    z + i = + or - sqrt(3-4i)
    It's given that sqrt(3-4i) = 2 - i
    so z = -i + or - (2 - i)
    so the 2 roots are z1 = 2(1-i) and z2 = -2
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    yup take the root of both sides as you're given the z^2 = 3-4i and that z=2-i


    so if you root both sides,

    z+i = +/-(2-i)

    so

    z+i = 2 - i
    z+i = -2 + i

    z = 2 - 2i or -2

    which are easy to draw on argand diagram
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    if youre given z = some sort of fraction, is there an easier way to find |z| and argz than having to rationalise ?
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    (Original post by kikzen)
    if youre given z = some sort of fraction, is there an easier way to find |z| and argz than having to rationalise ?
    |z/w| = |z|/|w| and arg(z/w) = argz - argw
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    (Original post by Bezza)
    |z/w| = |z|/|w| and arg(z/w) = argz - argw
    also true for multiplication
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    (Original post by keisiuho)
    also true for multiplication
    Except with a * and + rather than / and -
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    Is it also true for powers then?

    eg.

    does arg(|z|)^|w| = arg |z| * arg |w|?
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    (Original post by mik1a)
    Is it also true for powers then?

    eg.

    does arg(|z|)^|w| = arg |z| * arg |w|?
    I'm not sure why you have all the moduli signs in there - |z| would be a real number so arg|z| would be 0 or pi.

    |zw| = |z||w| and arg(zw) = argz + argw

    arg(z^n) = arg(z*z*z*z*z*.....*z) = n*argz
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    (Original post by Bezza)
    I'm not sure why you have all the moduli signs in there - |z| would be a real number so arg|z| would be 0 or pi.

    |zw| = |z||w| and arg(zw) = argz + argw

    arg(z^n) = arg(z*z*z*z*z*.....*z) = n*argz
    arg(z) = arg(z^(1/n)*z^(1/n)*z^(1/n)*z^(1/n)*...*z^(1/n)) = n*arg(z^(1/n))
    so arg(z^(1/n)) = 1/n arg(z)
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    so uh just like logs then eh
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    (Original post by kikzen)
    so uh just like logs then eh
    That's what I was thinking when I asked that question... although now that I think about it,

    (log a)^b =/= (log a)(log b)

    Silly me.
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    wot does arg mean?
 
 
 
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