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Complex numbers help

Hi, I've got to evaluate (root3 - i)^4 divided by (-1 + i)^3, I've so far gotten to (-2 - 2root3) + (2 - 2root3)i, but I have to convert it to exponential form, and the question is non-calculator.

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If your answer is supposed to be in the form reiθr e^{i\theta}, then you need to convert 3−i\sqrt{3} - i and −1+i-1 + i into this form and then use DeMoivre. It's not difficult - they are all "standard" angles.
Original post by jordanwu
Hi, I've got to evaluate (root3 - i)^4 divided by (-1 + i)^3, I've so far gotten to (-2 - 2root3) + (2 - 2root3)i, but I have to convert it to exponential form, and the question is non-calculator.

Draw a diagram and convert each complex number to the form reiθre^{i\theta}. That makes applying the powers, and dividing, a lot easier.
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Original post by DFranklin
If your answer is supposed to be in the form reiθr e^{i\theta}, then you need to convert 3−i\sqrt{3} - i and −1+i-1 + i into this form and then use DeMoivre. It's not difficult - they are all "standard" angles.


Ok, but I'm not sure if I've got the right exponential forms: 2e^i11/6pi and root2e^i3/4pi?
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Original post by jordanwu
Ok, but I'm not sure if I've got the right exponential forms: 2e^i11/6pi and root2e^i3/4pi?


Yeah those are correct
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Original post by DFranklin
If your answer is supposed to be in the form reiθr e^{i\theta}, then you need to convert 3−i\sqrt{3} - i and −1+i-1 + i into this form and then use DeMoivre. It's not difficult - they are all "standard" angles.


Not too sure about how to use that on the exponential form
Original post by jordanwu
Not too sure about how to use that on the exponential form


Well, if 3−i=2e−16πi\sqrt{3}-i=2e^{-\frac{1}{6}\pi i} as you found, then (3−i)4=(\sqrt{3}-i)^4=....?
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Original post by RDKGames
Well, if 3−i=2e−16πi\sqrt{3}-i=2e^{-\frac{1}{6}\pi i} as you found, then (3−i)4=(\sqrt{3}-i)^4=....?


Oh it's supposed to be -11/6pi? I've got 11/6pi
Original post by jordanwu
Oh it's supposed to be -11/6pi? I've got 11/6pi


if you have w = reiΘ

then wk = rkeikΘ
Original post by jordanwu
Oh it's supposed to be -11/6pi? I've got 11/6pi


e116πi=e−16πie^{\frac{11}{6}\pi i}=e^{-\frac{1}{6}\pi i}
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Original post by RDKGames
e116πi=e−16πie^{\frac{11}{6}\pi i}=e^{-\frac{1}{6}\pi i}


Anywhere I can look that rule up? I've definitely forgotten it. So I have to do 2^4 and (e^-1/6pi*i)^4? And same with the exponential on the denominator?
Original post by jordanwu
Anywhere I can look that rule up? I've definitely forgotten it. So I have to do 2^4 and (e^-1/6pi*i)^4? And same with the exponential on the denominator?

You can remove (or add) any number of integer multiples of 2π2\pi, as it's removing complete revolutions from the angle.
Original post by jordanwu
Anywhere I can look that rule up? I've definitely forgotten it. So I have to do 2^4 and (e^-1/6pi*i)^4? And same with the exponential on the denominator?


An exponential form complex number eiθe^{i \theta} repeats itself every 2π2\pi so eiθ=ei(2πk+θ)e^{i \theta}=e^{i (2\pi k + \theta)} for n∈Zn \in \mathbb{Z}

Pick θ=116π\theta= \frac{11}{6} \pi and k=−1k=-1 and you arrive at the result I've shown. However, it is much more intuitive to mark down the complex number 3−i\sqrt{3}-i on the complex plane, draw a line to it (its modulus), then notice how your angle is simply measured anticlockwise from R>0\mathbb{R}_{>0} while my angle is just measuring backwards (clockwise, hence the negative) to get my solution.


Anywho, yes you simply exponentiate these numbers to 4, then repeat similarly for the other complex number. Then you're simply in a position where you have reiθReiϕ\displaystyle \frac{re^{i \theta}}{Re^{i \phi}} which should be straight-forward to evaluate.
(edited 6 years ago)
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Original post by RDKGames
An exponential form complex number eiθe^{i \theta} repeats itself every 2π2\pi so eiθ=ei(2πk+θ)e^{i \theta}=e^{i (2\pi k + \theta)} for n∈Zn \in \mathbb{Z}

Pick θ=116π\theta= \frac{11}{6} \pi and k=−1k=-1 and you arrive at the result I've shown. However, it is much more intuitive to mark down the complex number 3−i\sqrt{3}-i on the complex plane, draw a line to it (its modulus), then notice how your angle is simply measured anticlockwise from R>0\mathbb{R}_{>0} while my angle is just measuring backwards (clockwise, hence the negative) to get my solution.


Anywho, yes you simply exponentiate these numbers to 4, then repeat similarly for the other complex number. Then you're simply in a position where you have reiθReiϕ\displaystyle \frac{re^{i \theta}}{Re^{i \phi}} which should be straight-forward to evaluate.


I'm not sure how to do (e^(11/6pi)i)^4
Original post by jordanwu
I'm not sure how to do (e^(11/6pi)i)^4
(ea)b=eab(e^a)^b = e^{ab}...
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Original post by DFranklin
(ea)b=eab(e^a)^b = e^{ab}...


e^(22/3*pi^4*i^4)?
Original post by jordanwu
e^(22/3*pi^4*i^4)?


Ah dear, someone's forgot a GCSE rule.

You have z=(3−i)4=(2e116πi)4z=(\sqrt{3}-i)^4=(2e^{\frac{11}{6}\pi i})^4

So... since (ea)b=eab(e^a)^b=e^{ab}, you simply have z=24e446πiz=2^4e^{\frac{44}{6}\pi i}
(edited 6 years ago)
Original post by jordanwu
e^(22/3*pi^4*i^4)?
No. Compare (e(11/6π)i)4(e^{(11/6\pi)i})^4 with the form (ea)b(e^a)^b.

What is a here?
What is b here?

So, then what is ab?

So what is eabe^{ab}?
Reply 18
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Original post by RDKGames
Ah dear, someone's forgot a GCSE rule.

You have z=(3−i)4=(e116πi)4z=(\sqrt{3}-i)^4=(e^{\frac{11}{6}\pi i})^4

So... since (ea)b=eab(e^a)^b=e^{ab}, you simply have z=e446πiz=e^{\frac{44}{6}\pi i}


Wow, what's wrong with me... don't even know how I got that. Ok, 16e^44/6pi*i / 2root2e^9/4pi*i?
Original post by jordanwu
Wow, what's wrong with me... don't even know how I got that. Ok, 16e^44/6pi*i / 2root2e^9/4pi*i?


Yeah

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