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    I need to know how to do this in matlab so i can compare the answer to my written work.
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    Using moivre's theorem?
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    With pen and paper please help me solve this.
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    (Original post by MatthewTG)
    With pen and paper please help me solve this.
    Saw your post about this concerning MATLAB, and I'm not quite sure how to check it on there either, but you can easily use something like Wolfram Alpha for verification.

    Anyhow, you need to express 1+j in the form \displaystyle re^{i\theta + 2k \pi i} where [rex]r[/tex] is the modulus and \theta is the argument of the complex number. This is for k\in \{0,1,2,3,4 \} and then take the 5th root of both sides.
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    (Original post by RDKGames)
    Saw your post about this concerning MATLAB, and I'm not quite sure how to check it on there either, but you can easily use something like Wolfram Alpha for verification.

    Anyhow, you need to express 1+j in the form \displaystyle re^{i\theta + 2k \pi i} where [rex]r[/tex] is the modulus and \theta is the argument of the complex number. This is for k\in \{0,1,2,3,4 \} and then take the 5th root of both sides.
    I do not understand what that re thing is
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    (Original post by MatthewTG)
    I do not understand what that re thing is
    Have you been taught different forms of a complex number??

    A complex number x+y\mathbf{i} such that x,y \in \mathbb{R} can be represented in the form re^{i\theta} where r=\sqrt{x^2+y^2} and \theta = arctan(\frac{y}{x})
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    (Original post by RDKGames)
    Have you been taught different forms of a complex number??

    A complex number x+y\mathbf{i} such that x,y \in \mathbb{R} can be represented in the polar form re^{i\theta} where [rex]r=\sqrt{x^2+y^2}[/tex] and \theta = arctan(\frac{y}{x})
    Only had like one or two lessons on that recently
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    (Original post by MatthewTG)
    Only had like one or two lessons on that recently
    Then you need to use this form.

    Alternatively, if you have been through De Moivre's Theorem, you can represent your complex number in the form z^5=r[\cos(\alpha)+\mathbf{i}\sin( \alpha)] where \alpha = \theta + 2\pi k so then z=\sqrt[5]{r}[\cos(\frac{1}{5} \alpha)+\sin(\frac{1}{5}\alpha) \mathbf{i} ]
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    (Original post by RDKGames)
    Then you need to use this form.

    Alternatively, if you have been through De Moivre's Theorem, you can represent your complex number in the form z^5=r[\cos(\alpha)+\mathbf{i}\sin( \alpha)] where \alpha = \theta + 2\pi k so then z=\sqrt[5]{r}[\cos(\frac{1}{5} \alpha)+\sin(\frac{1}{5}\alpha) \mathbf{i} ]
    ok thank you.
 
 
 
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