Complex numbers question
- Thread starter Louis1485
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D
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Use:Hi guys,
Can anyone help me with this?
What is each of the following numbers in the form a + bi? (e^9i + e^-9i) /2i, and 2e^(8+ (iπ/6)).
Thank you!
e^(i*Θ) = cos(Θ) + i * sin(Θ)
Please show us what you have tried and exactly where you are stuck.
Please follow the rules of posting in this forum, as enunciated at:
Please share your work/thoughts about this problem.
pka
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In addition to your not following the posting guidelines, the questions are not clear.
e^9i could mean \(e^{9i}\) or else \(e^{9}i\)
Now suspect that the question is \(\large{\dfrac{e^{9i}+e^{-9i}}{2i}}\).
Is that correct? If not please tell us what is correct!
Sorry, I'm posting on behalf of my girlfriend so I don't have her working. She has completed similar questions but was stuck on these two.
D
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pka
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With that change, the problem is made trivial.
If \(t\in\Re\) then \(e^{ti}=\cos(t)+i\sin(t)\), moreover:
cosine is an even function, \(\cos(-t)=\cos(t)\) & sine is an odd function, \(\sin(-t)=-\sin(t)\).
Please have your friend post her working.
ISTER_REG
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I don't care who writes here, I'd rather contribute something to the solution of the problem.
So: You have the following [MATH]\frac{e^{9i}-e^{-9i}}{2i}[/MATH]. Your question was about how to convert this form (Euler form [MATH]e^{i\phi}[/MATH]) into a Cartesian form ([MATH]a+ib[/MATH]). I see here two ways, once the manual one and once the rather "easier" one.
The simple way:
If you know this formula here [MATH]\sin(\phi) = \frac{e^{i\phi}-e^{-i\phi}}{2i}[/MATH], whose derivation is not that hard, then you can use this property "almost" directly. For your example:
[MATH]\frac{e^{9i}-e^{-9i}}{2i} = \sin(9) \approx 0.4121[/MATH]
The manual way:
For this you actually only have to use the Euler formula, which has been mentioned here many times, this is defined as [MATH]e^{\pm i\phi} = \cos(\phi) \pm i\sin(\phi)[/MATH]. Again, you must be careful to transform the fraction if necessary by expanding with the conjugate complex. For your example:
[MATH]e^{9i} =\cos(9) + i\sin(9)[/MATH][MATH]e^{-9i} = \cos(9) - i\sin(9)[/MATH]
[MATH]\frac{e^{9i}-e^{-9i}}{2i} = \frac{(\cos(9) + i\sin(9)) - (\cos(9) - i\sin(9))}{2i} = \frac{2i\sin(9)}{2i} = \sin(9) \approx 0.4121[/MATH]
So you see the imaginary part disappears here and the real part remains.
So: You have the following [MATH]\frac{e^{9i}-e^{-9i}}{2i}[/MATH]. Your question was about how to convert this form (Euler form [MATH]e^{i\phi}[/MATH]) into a Cartesian form ([MATH]a+ib[/MATH]). I see here two ways, once the manual one and once the rather "easier" one.
The simple way:
If you know this formula here [MATH]\sin(\phi) = \frac{e^{i\phi}-e^{-i\phi}}{2i}[/MATH], whose derivation is not that hard, then you can use this property "almost" directly. For your example:
[MATH]\frac{e^{9i}-e^{-9i}}{2i} = \sin(9) \approx 0.4121[/MATH]
The manual way:
For this you actually only have to use the Euler formula, which has been mentioned here many times, this is defined as [MATH]e^{\pm i\phi} = \cos(\phi) \pm i\sin(\phi)[/MATH]. Again, you must be careful to transform the fraction if necessary by expanding with the conjugate complex. For your example:
[MATH]e^{9i} =\cos(9) + i\sin(9)[/MATH][MATH]e^{-9i} = \cos(9) - i\sin(9)[/MATH]
[MATH]\frac{e^{9i}-e^{-9i}}{2i} = \frac{(\cos(9) + i\sin(9)) - (\cos(9) - i\sin(9))}{2i} = \frac{2i\sin(9)}{2i} = \sin(9) \approx 0.4121[/MATH]
So you see the imaginary part disappears here and the real part remains.
Last edited:
ISTER_REG, I reported your post #10 (or at least what was post # 10). You did not follow posting guidelines.
The OP in this thread had been given prompts about the posting guidelines himself to be clear on the
problem, who was working on the problem, and to get that person to show some work here so that help may
be given. That last part was not done. Then you ignored the above and posted solutions.
You are to not post solutions when there has been no attempts shown and helpers are in the middle of giving prompts.
ISTER_REG
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I understand what you are trying to say. The questioner asked a question on a whim where certain uncertainties were there. At worst, I wasted time formulating a flawless solution. I have hope that the questioner or his girlfriend here can learn something from this solution...