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Math Help - Complex number calculation and expresing it in polar form

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    Post Complex number calculation and expresing it in polar form

    Calculate (1-j)^5 and express the answer in polar form. 'j' being the imaginary number.
    If you could also show me the working out i'd appreciate it.
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    Re: Complex number calculation and expresing it in polar form

    Quote Originally Posted by calmath View Post
    Calculate (1-j)^5 and express the answer in polar form. 'j' being the imaginary number.
    If you could also show me the working out
    That is not the way it is done here.
    You show some work. Tell us what you don't understand.
    We try to help you then.

    Here is a start: 1-i=\sqrt{2}\exp\left(\frac{-\pi}{4}\right).
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    Re: Complex number calculation and expresing it in polar form

    Quote Originally Posted by Plato View Post
    That is not the way it is done here.
    You show some work. Tell us what you don't understand.
    We try to help you then.

    Here is a start: 1-i=\sqrt{2}\exp\left(\frac{-\pi}{4}\right).
    I got up to the following part;

    Where it says calculate I didnt quite understand what it wants so id did the following;

    (1-i)^5=1-i-i^5=2-i

    Modulus is;
    Sqrt(2^2+1^2 = Sqrt(5)

    Argument(2-i)= atan(-1/2)= -26.6
    (atan(-1/2))+360= 333.4degrees

    In polar form, 2-i is: Sqrt(5), theta= 333.4
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    Re: Complex number calculation and expresing it in polar form

    Quote Originally Posted by calmath View Post
    I got up to the following part;

    Where it says calculate I didnt quite understand what it wants so id did the following;

    (1-i)^5=1-i-i^5=2-i
    Wrong. If you were going to expand these brackets, you should have (1 - i)(1 - i)(1 - i)(1 - i)(1 - i), which should give you a LOT more than 1 - i - i^5.

    As you have been told already, convert to the polar form which Plato did in post 2, then use DeMoivre's Theorem, like you have been instructed.
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    MHF Contributor FernandoRevilla's Avatar
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    Re: Complex number calculation and expresing it in polar form

    Out of curiosity: for a complex number z=x+yj with |x|=|y| the computation of (x\pm yj)^n in binomial form can be simplified taking into account that (x\pm yj)^2 is a pure imaginary number. In our case

    (1-j)^5=[(1-j)^2]^2(1-j)=(-2j)^2(1-j)=-4(1-j)=-4+4j .
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