September 21, 2020, 01:46:08 pm

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##### Re: Specialist 1/2 Question Thread!
« Reply #360 on: August 13, 2020, 05:34:33 pm »
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What are you struggling with? What have you tried yourself?

I tried multiplying it repeatedly, utilising the fact that i^2 is a real number. Can't get more minute than that, need help

#### keltingmeith

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##### Re: Specialist 1/2 Question Thread!
« Reply #361 on: August 13, 2020, 05:37:44 pm »
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I tried multiplying it repeatedly, utilising the fact that i^2 is a real number. Can't get more minute than that, need help

Yeah, so as you've seen, that'll quickly become an issue because it only works for when n is an integer - which it might not be. Have you learned about using De Moivre's theorem? Because that'll be integral to solving this question
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##### Re: Specialist 1/2 Question Thread!
« Reply #362 on: August 13, 2020, 05:39:58 pm »
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Yeah, so as you've seen, that'll quickly become an issue because it only works for when n is an integer - which it might not be. Have you learned about using De Moivre's theorem? Because that'll be integral to solving this question

I haven't learn that, no. I'll learn out the theorem then try the problem again, if that doesn't work then I'll come back here. Thanks for telling me about De Moivre's theorem!

#### keltingmeith

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##### Re: Specialist 1/2 Question Thread!
« Reply #363 on: August 13, 2020, 05:43:11 pm »
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I haven't learn that, no. I'll learn out the theorem then try the problem again, if that doesn't work then I'll come back here. Thanks for telling me about De Moivre's theorem!

All good - information on the internet can be confusing, but there should be information in your textbook - and here's a quick summary:

Once you've converted your complex number, z, to polar form, then it has a modulus (r) and an argument (theta), and has the form:

$z=r\times \text{cis}(\theta)$

Then, the form of z^n is:

$z^n=\left[r\times\text{cis}(\theta)\right]^n=r^n\times\text{cis}(n\theta)$
Currently Undertaking: Doctor of Philosophy (PhD) in Supramolecular Photochemistry (things that don't bond but they do and glow pretty colours)

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##### Re: Specialist 1/2 Question Thread!
« Reply #364 on: August 13, 2020, 07:31:56 pm »
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All good - information on the internet can be confusing, but there should be information in your textbook - and here's a quick summary:

Once you've converted your complex number, z, to polar form, then it has a modulus (r) and an argument (theta), and has the form:

$z=r\times \text{cis}(\theta)$

Then, the form of z^n is:

$z^n=\left[r\times\text{cis}(\theta)\right]^n=r^n\times\text{cis}(n\theta)$

Hi, I tried using DM's theorem. I converted the given problem to polar form, and expanded n. My approach has been that basically sin((-pi*n)/3 cannot equal i, as if it did then i times i means a real number. However, I tried to use the CAS to solve for n, but it's simply returning "false". Where to now?

#### S_R_K

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##### Re: Specialist 1/2 Question Thread!
« Reply #365 on: August 13, 2020, 08:38:29 pm »
+1
Hi, I tried using DM's theorem. I converted the given problem to polar form, and expanded n. My approach has been that basically sin((-pi*n)/3 cannot equal i, as if it did then i times i means a real number. However, I tried to use the CAS to solve for n, but it's simply returning "false". Where to now?

The argument of an imaginary number is $\frac{\pi}{2} + \pi k$ where k is an integer. Equate that to the argument of $(2-2\sqrt{3}i)^n$ and solve for the smallest positive integer n.

#### M-D

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##### Re: Specialist 1/2 Question Thread!
« Reply #366 on: August 30, 2020, 08:15:30 pm »
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Hi,
I need help with the following question:

$a=2i - 3j + k, b=2i - 4j + 5k, c= -i - 4j + 2k$ Find the values of p and q such that $a + pb + qc$ is parallel to the x-axis.

I have worked out that:

$a + pb + qc = i(2+2p-q) + j(-3-4p-4q) + k(1+5p+2q)$

I don't know how to get specific values for p and q so $a + pb + qc$ is parallel to the x-axis.

Any assistance will be much appreciated.

#### keltingmeith

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##### Re: Specialist 1/2 Question Thread!
« Reply #367 on: August 30, 2020, 08:19:21 pm »
+1
Hi,
I need help with the following question:

$a=2i - 3j + k, b=2i - 4j + 5k, c= -i - 4j + 2k$ Find the values of p and q such that $a + pb + qc$ is parallel to the x-axis.

I have worked out that:

$a + pb + qc = i(2+2p-q) + j(-3-4p-4q) + k(1+5p+2q)$

I don't know how to get specific values for p and q so $a + pb + qc$ is parallel to the x-axis.

Any assistance will be much appreciated.

Well, if the vector is parallel to the x-axis, then it should be equal to some multiple of the i unit vector - does that help?
Currently Undertaking: Doctor of Philosophy (PhD) in Supramolecular Photochemistry (things that don't bond but they do and glow pretty colours)

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Bachelor of Science Advanced (Research) - Monash University, majoring in Mathematical Statistics and Chemistry
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