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    a) Given that z = 2 - i, show that z^2 = 3 - 4i.
    b) Hence, or otherwise, find the roots, z1 and z2, of the equation (z+i)^2 = 3 - 4i.
    I got part a right, but can someone teach me how i should start on part b please?
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    If (z+i)^2= 3-4i then z+i =the roots from part a


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    From a we know z+i = sqrt(3-4i) so z+i=+ or - sqrt(3-4i) and hence z+i=+ or -(2-i) which gives you z=2-2i and z=-2

    Hope it helps!
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    you can do part (a) and you cannot do part (b)? ....
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    (Original post by TeeEm)
    you can do part (a) and you cannot do part (b)? ....
    yes.
    for part b, i started with (z-i)^2 = 3 - 4i
    2-i = + or - sqrt(3-4i)
    What should i do next? as i dont really get the question. When it said find the roots, does it mean find the square root of 3-4i?
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    (Original post by alesha98)
    yes.
    for part b, i started with (z-i)^2 = 3 - 4i
    2-i = + or - sqrt(3-4i)
    What should i do next? as i dont really get the question. When it said find the roots, does it mean find the square root of 3-4i?
    teaching till 23.15

    the others above are giving you the correct method
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    (Original post by TeeEm)
    teaching till 23.15

    the others above are giving you the correct method
    sorry i am learning it myself and i just dont get this type of question, when it asks for the roots. in the case of part b, how does it show 2 - i is the root?
    i am happy to stay till 23,15 to learn FP1, thankyou
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    (Original post by TeeEm)
    teaching till 23.15

    the others above are giving you the correct method
    hi can someone check my working? i managed to work out the solution

    Since from part a, i can get
    z^2 = 3 - 4i
    z = +or- sqrt (3-4i)
    2-i = +or- sqrt (3-4i)
    sqrt(3-4i) = 2-i or -2+i

    so in part b,
    (z+i)^2 = 3 - 4i
    z+i = +or- sqrt (3-4i)
    z+i = 2-i or -2+i
    z = 2-2i or -2
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    (Original post by alesha98)
    hi can someone check my working? i managed to work out the solution

    Since from part a, i can get
    z^2 = 3 - 4i
    z = +or- sqrt (3-4i)
    2-i = +or- sqrt (3-4i)
    sqrt(3-4i) = 2-i or -2+i

    so in part b,
    (z+i)^2 = 3 - 4i
    z+i = +or- sqrt (3-4i)
    z+i = 2-i or -2+i
    z = 2-2i or -2
    if z = 2-i
    z^2 = (2-i)^2

    Second part is easy, put some thought into it
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    (Original post by LelouchViRuge)
    if z = 2-i
    z^2 = (2-i)^2

    Second part is easy, put some thought into it
    (2-i)^2 = (2-i)(2-i), i got this right at the begining
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    (Original post by alesha98)
    sorry i am learning it myself and i just dont get this type of question, when it asks for the roots. in the case of part b, how does it show 2 - i is the root?
    i am happy to stay till 23,15 to learn FP1, thankyou
    I will write a similar question with a full solution to enlighten you, tomorrow (day off)
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    (Original post by TeeEm)
    I will write a similar question with a full solution to enlighten you, tomorrow (day off)
    ok thank you so much
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    (Original post by alesha98)
    ok thank you so much
    let me scan an example I made earlier ...
    different approaches
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    (Original post by alesha98)
    ok thank you so much
    ..
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    (Original post by alesha98)
    i dont really get the question. When it said find the roots, does it mean find the square root of 3-4i?
    I think what you were confused about is what a root is? Root just means solution,so it's asking for z, in case you hadn't worked that out yet
 
 
 
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