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Year 12s: what will be the first way you research your future options? >>

Exponential Complex Number

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For 11c I've managed to get z^2=r^2 e^2i0 however I cannot get the answer of (2cos0)e^i0

Thank you :smile:

:smile:

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(edited 9 years ago)

Original post by Mutleybm1996

ImageUploadedByStudent Room1417201605.178022.jpg

For 11c I've managed to get z^2=r^2 e^2i0 however I cannot get the answer of (2cos0)e^i0

Thank you :smile:

:smile:

Posted from TSR Mobile

Factorise

e^{i\theta}

out of

e^{2i\theta} + 1

OP

Original post by atsruser

X

So:

e^{i\theta}

out of

e^{i\theta} (e^{i\theta} + 1/{e^{i\theta}})

Posted from TSR Mobile

(edited 9 years ago)

Original post by Mutleybm1996

ImageUploadedByStudent Room1417201605.178022.jpg

For 11c I've managed to get z^2=r^2 e^2i0 however I cannot get the answer of (2cos0)e^i0

Thank you :smile:

:smile:

Posted from TSR Mobile

Well r is just 1, and then use De-Moivre.

OP

Original post by joostan

Well r is just 1, and then use De-Moivre.

We haven't covered De-Moivre's

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Original post by Mutleybm1996

We haven't covered De-Moivre's

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No worries, I assume you have covered Euler's Formula, you can simply expand the square from there.

OP

Original post by joostan

No worries, I assume you have covered Euler's Formula, you can simply expand the square from there.

ImageUploadedByStudent Room1417203476.789229.jpg

Like this?

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Original post by Mutleybm1996

ImageUploadedByStudent Room1417203476.789229.jpg

Like this?

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yup then using some trig identities you're done

(edited 9 years ago)

Original post by Mutleybm1996

So:

e^{i\theta}

out of

e^{i\theta} (e^{i\theta} + 1/{e^{i\theta}})

Posted from TSR Mobile

Correct, but I'd write

e^{2i\theta} +1 = e^{i\theta}(e^{i\theta} + e^{-i\theta})

since then

e^{-i\theta} = \cos (-\theta) + i \sin(-\theta)

gives you the result that you want, if you know how sin/cos behave for negative arguments.

OP

Original post by joostan

yup then using some trig identities you're done

ImageUploadedByStudent Room1417203758.348588.jpg

Sorry, it's been a while since I did C3/4

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Original post by Mutleybm1996

ImageUploadedByStudent Room1417203758.348588.jpg

Sorry, it's been a while since I did C3/4

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Combine

\cos(2x)=\cos^2(x)-sin^(x)

\sin^(x)=1-\cos^2(x)

to get an identity for:

\cos(2x)

involving only cos terms

OP

Original post by atsruser

Correct, but I'd write

e^{2i\theta} +1 = e^{i\theta}(e^{i\theta} + e^{-i\theta})

since then

e^{-i\theta} = \cos (-\theta) + i \sin(-\theta)

gives you the result that you want, if you know how sin/cos behave for negative arguments.

Thank you! I hadn't thought about writing it as e^-i(theta) to get the 1!

Thanks :smile:

:smile:

that worked out very nicely :smile:

:smile:

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OP

Original post by joostan

Combine

\cos(2x)=\cos^2(x)-sin^(x)

\sin^(x)=1-\cos^2(x)

to get an identity for:

\cos(2x)

involving only cos terms

I tried that earlier and didn't get anywhere, I got the cos(2x)= 2cos^2x -1 but I couldn't link that to the answer

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Original post by Mutleybm1996

I tried that earlier and didn't get anywhere, I got the cos(2x)= 2cos^2x -1 but I couldn't link that to the answer

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Ah, I imagine the problem may have been forgetting the 1 from the equation, but I see you've done it now :smile:

:smile:

OP

Original post by atsruser

Correct, but I'd write

e^{2i\theta} +1 = e^{i\theta}(e^{i\theta} + e^{-i\theta})

since then

e^{-i\theta} = \cos (-\theta) + i \sin(-\theta)

gives you the result that you want, if you know how sin/cos behave for negative arguments.

If I could pester you once more, for part d I got e^i0(-2sin0) I assume that is the same as e^i(0-pi/2)(2sin0)...correct?

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OP

Original post by joostan

Combine

\cos(2x)=\cos^2(x)-sin^(x)

\sin^(x)=1-\cos^2(x)

to get an identity for:

\cos(2x)

involving only cos terms

ImageUploadedByStudent Room1417205096.429361.jpg

One final question if that's okay, how would I go about doing 13c? I have successfully managed a and b

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Original post by Mutleybm1996

ImageUploadedByStudent Room1417205096.429361.jpg

One final question if that's okay, how would I go about doing 13c? I have successfully managed a and b

Posted from TSR Mobile

You have a First Order seperable ODE.
You can treat

i

as a constant and then solve as you would any other d.e.

Original post by Mutleybm1996

If I could pester you once more, for part d I got e^i0(-2sin0) I assume that is the same as e^i(0-pi/2)(2sin0)...correct?

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It would be easier to read what you are writing if you use latex.

That doesn't look correct.

1-z^2 = 1-e^{2i\theta} = (e^{-i\theta}-e^{i\theta})e^{i\theta} = \\ (\cos\theta - i\sin\theta -(\cos\theta + i\sin\theta))e^{i\theta} = \\ -2i\sin\theta e^{i\theta} = 2\sin\theta e^{-i\pi/2} e^{i\theta} = 2\sin\theta e^{i(\theta-\pi/2)}

so you are missing a factor of

i

OP

Original post by atsruser

It would be easier to read what you are writing if you use latex.

That doesn't look correct.

1-z^2 = 1-e^{2i\theta} = (e^{-i\theta}-e^{i\theta})e^{i\theta} = \\ (\cos\theta - i\sin\theta -(\cos\theta + i\sin\theta))e^{i\theta} = \\ -2i\sin\theta e^{i\theta} = 2\sin\theta e^{-i\pi/2} e^{i\theta} = 2\sin\theta e^{i(\theta-\pi/2)}

so you are missing a factor of

i

Thank you

:smile:

sorry, I've been meaning to learn how to use it, but I've never quite gotten around to it. May I ask how you got from the penultimate stage to your final stage?

Thanks :smile:

:smile:

+rep to both of you :smile:

:smile:

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Original post by Mutleybm1996

May I ask how you got from the penultimate stage to your final stage?

Note that

-i

is on the negative y-axis at (0,-1) so it has a length of 1 and an argument of

-\pi/2

and thus in exponential form

-i=e^{-i\pi/2}

.

Then the final step merely uses the fact that

a^m \times a^n = a^{m+n}

also holds when

m,n

are complex; I'm just adding the powers of the complex exponentials together, and factorising out the

i

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