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how to solve z^5=1+j in matlab?

I need to know how to do this in matlab so i can compare the answer to my written work.
Reply 1
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3121
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Using moivre's theorem?
Reply 2
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MatthewTG
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With pen and paper please help me solve this.
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+j1+j in the form reiθ+2kπi\displaystyle re^{i\theta + 2k \pi i} where [rex]r is the modulus and θ\theta is the argument of the complex number. This is for k∈{0,1,2,3,4}k\in \{0,1,2,3,4 \} and then take the 5th root of both sides.
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MatthewTG
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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+j1+j in the form reiθ+2kπi\displaystyle re^{i\theta + 2k \pi i} where [rex]r
is the modulus and θ\theta is the argument of the complex number. This is for k∈{0,1,2,3,4}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
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+yix+y\mathbf{i} such that x,y∈Rx,y \in \mathbb{R} can be represented in the form reiθre^{i\theta} where r=x2+y2r=\sqrt{x^2+y^2} and θ=arctan(yx)\theta = arctan(\frac{y}{x})
(edited 7 years ago)
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MatthewTG
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Original post by RDKGames
Have you been taught different forms of a complex number??

A complex number x+yix+y\mathbf{i} such that x,y∈Rx,y \in \mathbb{R} can be represented in the polar form reiθre^{i\theta} where [rex]r=\sqrt{x^2+y^2}
and θ=arctan(yx)\theta = arctan(\frac{y}{x})

Only had like one or two lessons on that recently
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 z5=r[cos⁡(α)+isin⁡(α)]z^5=r[\cos(\alpha)+\mathbf{i}\sin( \alpha)] where α=θ+2πk\alpha = \theta + 2\pi k so then z=r5[cos⁡(15α)+sin⁡(15α)i]z=\sqrt[5]{r}[\cos(\frac{1}{5} \alpha)+\sin(\frac{1}{5}\alpha) \mathbf{i} ]
Reply 8
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MatthewTG
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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 z5=r[cos⁡(α)+isin⁡(α)]z^5=r[\cos(\alpha)+\mathbf{i}\sin( \alpha)] where α=θ+2πk\alpha = \theta + 2\pi k so then z=r5[cos⁡(15α)+sin⁡(15α)i]z=\sqrt[5]{r}[\cos(\frac{1}{5} \alpha)+\sin(\frac{1}{5}\alpha) \mathbf{i} ]


ok thank you.

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