Further Maths - Complex Numbers finding distance

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#1
Hello,
This is the question:

I have formed a function in z (a relative starting point) to which I can apply the function 4 times to see the complex number of the end position. Using z=0 allows me to find the complex number from origin whose modulus will yield me the total distance.
I have done this:

However, I’m unable to show how my distance manipulates into the form they ask for, any help is appreciated thanks
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4 months ago
#2
Remember that the modulus of the division of two complex numbers is the division of each modulus.
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#3
(Original post by mqb2766)
Remember that the modulus of the division of two complex numbers is the division of each modulus.
I understand, however, I’m not sure what to do with the additional 1 on the top and bottom
Attachment 925900
I’ve tried to do that however, I still don’t see how that manipulates into the desired form
Last edited by BrandonS15; 4 months ago
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4 months ago
#4
(Original post by BrandonS15)
I understand, however, I’m not sure what to do with the additional 1 on the top and bottom
Could try seeing
e^ix - 1 = e^(ix/2)( ....)
helps? Answer spotting, you want to half the angle on top and bottom.
Not worked it through though.

Last edited by mqb2766; 4 months ago
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4 months ago
#5
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4 months ago
#6
oops forgot to square the cos 2pi/9 -1
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4 months ago
#7
(Original post by Idg a damn)
...
The previous hint is easier. And the aim is to give hints, not solutions.
Last edited by mqb2766; 4 months ago
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4 months ago
#8
(Original post by mqb2766)
The previous hint is easier.
doesn't matter as long as it works
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4 months ago
#9
(Original post by mqb2766)
The previous hint is easier. And the aim is to give hints, not solutions.
Ah I get what you mean now ,should have thought of your method and use it instead
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4 months ago
#10
(Original post by mqb2766)
Could try seeing
e^ix - 1 = e^(ix/2)( ....)
helps? Answer spotting, you want to half the angle on top and bottom.
Not worked it through though.

From here , you can see that the real parts cancel out , and then...
Last edited by Idg a damn; 4 months ago
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4 months ago
#11
(Original post by Idg a damn)
Ah I get what you mean now ,should have thought of your method and use it instead
Yes, the sin half angle drops out of the conjugate expression.
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#12
(Original post by mqb2766)
Could try seeing
e^ix - 1 = e^(ix/2)( ....)
helps? Answer spotting, you want to half the angle on top and bottom.
Not worked it through though.

I’ve tried to apply the z^n + z^-n = 2cosntheta with manipulation here but I’ve found no luck, I’m still confused as to how that value of S, which is the distance moved by the ant with 4 iterations of the motion described, manipulates into the form they ask for
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4 months ago
#13
(Original post by BrandonS15)
I’ve tried to apply the z^n + z^-n = 2cosntheta with manipulation here but I’ve found no luck, I’m still confused as to how that value of S, which is the distance moved by the ant with 4 iterations of the motion described, manipulates into the form they ask for
don't think too far

it's like I said before: the real parts get canceled out

try and sub the polar forms for their complex forms and see what happens
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4 months ago
#14
(Original post by mqb2766)
Could try seeing
e^ix - 1 = e^(ix/2)( ....)
helps? Answer spotting, you want to half the angle on top and bottom.
Not worked it through though.

after you have done this that is
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4 months ago
#15
(Original post by BrandonS15)
I’ve tried to apply the z^n + z^-n = 2cosntheta with manipulation here but I’ve found no luck, I’m still confused as to how that value of S, which is the distance moved by the ant with 4 iterations of the motion described, manipulates into the form they ask for
Don't square.
Take the magnitude of numerator and denominator separately and apply the earlier formula on both terms.
Remember the magnitude of product is the product of magnitudes and the sin and complex conjugate identity.
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#16
(Original post by Idg a damn)
don't think too far

it's like I said before: the real parts get canceled out

try and sub the polar forms for their complex forms and see what happens

Okay I’ve made some progress however, I have the e^2pi(i)/3 complex number in the numerator which I don’t want to be there so where have I gone wrong?
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4 months ago
#17
(Original post by BrandonS15)

Okay I’ve made some progress however, I have the e^2pi(i)/3 complex number in the numerator which I don’t want to be there so where have I gone wrong?
You're not using
|e^i*| = 1
Or the other stuff? It's easier than you're making it if you use the magnitude correctly.

Remember z^2 is not the magnitude squared.
Last edited by mqb2766; 4 months ago
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#18
(Original post by mqb2766)
You're not using
|e^i*| = 1
Or the other stuff?

Is this a correct method? Sorry the solution bank squared both sides but I get how both methods work with that statement now
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4 months ago
#19
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4 months ago
#20
(Original post by BrandonS15)

Is this a correct method? Sorry the solution bank squared both sides but I get how both methods work with that statement now
Yes. If you square the magnitue, you multiply z by its conjugate.
That will get you to the answer, but if you're happy with this quoted solution, it's more elegant.
Last edited by mqb2766; 4 months ago
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