Maximus 190
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This is my first step using Euler’s formula Name:  64C5C729-AED1-4A94-ADF3-7E0A6F02DDB7.jpg.jpeg
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Not sure what do do next to get it into the standard modulus argument form of r(cos(a) isin(a)) which seems to be the right way to go (worked for Part A).
Would appreciate a little help! Perhaps there is an identity I could apply or something else?
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the bear
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so to find -z you can think of it as a rotation of 180 degrees around the origin ?
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the bear
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for part B consider the top of the fraction as 1 + 0i

when you divide complex numbers you divide the moduli ( here 1 / 4 ), and subtract the arguments.... here 0 - α
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Maximus 190
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(Original post by the bear)
so to find -z you can think of it as a rotation of 180 degrees around the origin ?
Okay. So surely shouldn’t 1/z give the same argument so it’s theta but that is incorrect?
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Maximus 190
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(Original post by the bear)
for part B consider the top of the fraction as 1 + 0i

when you divide complex numbers you divide the moduli ( here 1 / 4 ), and subtract the arguments.... here 0 - α
Oh okay that is cool! Thanks
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Maximus 190
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(Original post by the bear)
so to find -z you can think of it as a rotation of 180 degrees around the origin ?
What about the angle for Part C? Name:  F8711EC3-68B4-4C18-BE42-6AE887E9C869.jpg.jpeg
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The complex conjugate is just a reflection in the real axis so i*sin(a) becomes -i*sin(a) which equals i*sin(a pi) but it says that’s wrong? Name:  3CDDB8ED-18E1-4EF4-B74F-4DDA9134EF99.jpg.jpeg
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Last edited by Maximus 190; 1 month ago
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mqb2766
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(Original post by Maximus 190)
What about the angle for Part C? Name:  F8711EC3-68B4-4C18-BE42-6AE887E9C869.jpg.jpeg
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The complex conjugate is just a reflection in the real axis so i*sin(a) becomes -i*sin(a) which equals i*sin(a pi) but it says that’s wrong? Name:  3CDDB8ED-18E1-4EF4-B74F-4DDA9134EF99.jpg.jpeg
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The best way to view it is, as you say, a reflection in the x-axis. You've done a rotation by 180 degrees which is different. You want
a -> -a
Then map it to the range you want (positive angle, measured anticlockwise from the positive x-axis).

The sin bit is correct but you've forgotten to reason about the real part of z which is cos().
Last edited by mqb2766; 1 month ago
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RDKGames
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(Original post by Maximus 190)

Not sure what do do next to get it into the standard modulus argument form of r(cos(a) isin(a)) which seems to be the right way to go (worked for Part A).
Would appreciate a little help! Perhaps there is an identity I could apply or something else?
I know you already got this, but you could've made your life simpler by writing

\dfrac{1}{4e^{i \alpha}} = \dfrac{1}{4}e^{-i\alpha}
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Maximus 190
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(Original post by mqb2766)
The best way to view it is, as you say, a reflection in the x-axis. You've done a rotation by 180 degrees which is different. You want
a -> -a
Then map it to the range you want (positive angle, measured anticlockwise from the positive x-axis).

The sin bit is correct but you've forgotten to reason about the real part of z which is cos().
Oh yeah I forgot changing the sin angle will change the cos and hence the real part too :P got it now thanks

(Original post by RDKGames)
I know you already got this, but you could've made your life simpler by writing

\dfrac{1}{4e^{i \alpha}} = \dfrac{1}{4}e^{-i\alpha}
Oh yeah you’re right that is a bit simpler, thanks
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