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    (Original post by chemistryboi)
    another method that is more efficient
    Consider the arg of 1+i, and what happens when you to the arg when you raise the number to the power of 4.
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    (Original post by ghostwalker)
    Consider the arg of 1+i, and what happens when you to the arg when you raise the number to the power of 4.
    im not sure, i^4 is equal to 1. I am without a clue on this
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    (Original post by chemistryboi)
    im not sure, i^4 is equal to 1. I am without a clue on this
    Ok, have you come across "arg", or the format re^{i\theta} for a complex number?
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    (Original post by ghostwalker)
    Ok, have you come across "arg", or the format re^{i\theta} for a complex number?
    I have not come across that format, my understanding of the argument is the angle (usually called theta) in radians measured anticlockwise from y=0 to the line Z
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    (Original post by chemistryboi)
    I have not come across that format, my understanding of the argument is the angle (usually called theta) in radians measured anticlockwise from y=0 to the line Z
    :holmes: How about the \cos\theta +i\sin\theta format, and what happens when you raise it to a power?
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    (Original post by ghostwalker)
    :holmes: How about the \cos\theta +i\sin\theta format, and what happens when you raise it to a power?
    I don't know where I have gotten this idea from but does it rotate? As in change quadrant but with a different angle
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    (Original post by ghostwalker)
    Ok, have you come across "arg", or the format re^{i\theta} for a complex number?
    This is an FP2 formula.

    (Original post by chemistryboi)
    Hi TSR
    hence show that ((2+3i)/(5+i))^4 is real and determine its value.
    You know what (2+3i)/(5+i) simplifies to, just raise that to the power 4. I.E (lambda(1+i))^4 = (lambda)^4(1+i)^4
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    (Original post by chemistryboi)
    I don't know where I have gotten this idea from but does it rotate? As in change quadrant but with a different angle
    It does "rotate". When you multply two complex numbers together you sum their angles, so to speak.

    But since you've not come across these format yet (they are covered at A-level). I suggest substitute your expression \lambda (1+i), and square it, and square it again, to get the fourth power. After the first squaring, it will be a simpler format.
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    (Original post by ghostwalker)
    .
    Thank you for helping me out on this one. I greatly appreciate it.
 
 
 
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