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March 28, 2024, 08:11:10 pm

Author Topic: 4U Maths Question Thread  (Read 659775 times)  Share 

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louisaaa01

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Re: 4U Maths Question Thread
« Reply #2415 on: November 07, 2019, 08:30:08 pm »
+2
How would I solve this question? I know that if you multiply a complex number by i, you get a rotation of pi/2. I also know that arg(zw)=arg(z)+arg(w), but how would I use these to show the working out??

Any help is appreciated :)

Hi Mani.s_

Speaking in general terms, when you multiply a complex number by (cosθ + isinθ), you're really rotating it anticlockwise about the origin by θ radians. Indeed, a rotation by i will produce an anticlockwise rotation of π/2 since i in mod/arg form is really (cos(π/2) + isin(π/2)).

For this question, you are essentially given a value of θ. What you must do is convert cosθ + isinθ (the mod/arg form) into the Cartesian form - more generally expressed as x + iy. To do so, you use exact trigonometric values, e.g. for b, cos(π/3) + isin(π/3) becomes 1/2 + i√3/2.

I hope this helps!
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mani.s_

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Re: 4U Maths Question Thread
« Reply #2416 on: November 08, 2019, 11:05:48 pm »
0
Hi Mani.s_

Speaking in general terms, when you multiply a complex number by (cosθ + isinθ), you're really rotating it anticlockwise about the origin by θ radians. Indeed, a rotation by i will produce an anticlockwise rotation of π/2 since i in mod/arg form is really (cos(π/2) + isin(π/2)).

For this question, you are essentially given a value of θ. What you must do is convert cosθ + isinθ (the mod/arg form) into the Cartesian form - more generally expressed as x + iy. To do so, you use exact trigonometric values, e.g. for b, cos(π/3) + isin(π/3) becomes 1/2 + i√3/2.

I hope this helps!
Thank you so much!!!

mani.s_

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Re: 4U Maths Question Thread
« Reply #2417 on: November 08, 2019, 11:06:42 pm »
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Hi, can someone explain what fractals are???

Thanks :)

mani.s_

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Re: 4U Maths Question Thread
« Reply #2418 on: November 08, 2019, 11:17:05 pm »
0
Hi Mani.s_

Speaking in general terms, when you multiply a complex number by (cosθ + isinθ), you're really rotating it anticlockwise about the origin by θ radians. Indeed, a rotation by i will produce an anticlockwise rotation of π/2 since i in mod/arg form is really (cos(π/2) + isin(π/2)).

For this question, you are essentially given a value of θ. What you must do is convert cosθ + isinθ (the mod/arg form) into the Cartesian form - more generally expressed as x + iy. To do so, you use exact trigonometric values, e.g. for b, cos(π/3) + isin(π/3) becomes 1/2 + i√3/2.

I hope this helps!
Thank you so much!!!

shekhar.patel

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Re: 4U Maths Question Thread
« Reply #2419 on: December 18, 2019, 10:38:17 pm »
0
I need help in this question, and at the end the messed up letter is meant be be alpha conjucate

DrDusk

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Re: 4U Maths Question Thread
« Reply #2420 on: December 18, 2019, 11:17:40 pm »
0
I need help in this question, and at the end the messed up letter is meant be be alpha conjucate
« Last Edit: December 18, 2019, 11:24:40 pm by DrDusk »

RuiAce

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Re: 4U Maths Question Thread
« Reply #2421 on: December 18, 2019, 11:22:19 pm »
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Did you justify why the bar vanishes off the coefficients \(a\), \(b\) and \(c\)?

DrDusk

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Re: 4U Maths Question Thread
« Reply #2422 on: December 18, 2019, 11:23:31 pm »
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Did you justify why the bar vanishes off the coefficients \(a\), \(b\) and \(c\)?
I knew this was coming hahaha  ;)


shekhar.patel

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Re: 4U Maths Question Thread
« Reply #2423 on: December 19, 2019, 03:48:47 pm »
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Oh, I see. But how do we know that we need to sub alpha at the start. Thanks

RuiAce

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Re: 4U Maths Question Thread
« Reply #2424 on: December 19, 2019, 03:51:03 pm »
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Oh, I see. But how do we know that we need to sub alpha at the start. Thanks
From the fact that \(\alpha\) is by definition a root of \(ax^2+bx+c=0\).

You're not being asked to show that \(\alpha\) is a root; you're given it. (What has to be shown is that the same result holds when \(\overline{\alpha}\) is subbed in.)

shekhar.patel

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Re: 4U Maths Question Thread
« Reply #2425 on: December 19, 2019, 03:54:14 pm »
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From the fact that \(\alpha\) is by definition a root of \(ax^2+bx+c=0\).

You're not being asked to show that \(\alpha\) is a root; you're given it. (What has to be shown is that the same result holds when \(\overline{\alpha}\) is subbed in.)

Got it. Cheers

shekhar.patel

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Re: 4U Maths Question Thread
« Reply #2426 on: December 19, 2019, 04:06:51 pm »
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I have this question, and my attempted answer is below, I am not completely sure if it is the right approach.
Any assistance would be great.

DrDusk

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Re: 4U Maths Question Thread
« Reply #2427 on: December 19, 2019, 04:27:57 pm »
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I have this question, and my attempted answer is below, I am not completely sure if it is the right approach.
Any assistance would be great.
You can't do that because the conjugate doesn't apply to a, b and c IF they are real. We haven't proved this so you can't just use it in your proof like that. Instead you should do this:



fun_jirachi

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Re: 4U Maths Question Thread
« Reply #2428 on: December 19, 2019, 04:30:05 pm »
+1
Hey there!

Could you possibly explain your line of thinking when writing 'since the conjugate roots only applies to imaginary components we can say a, b and c are real?'. I'm a little confused - we could properly correct/understand/explain better if you clarified this :)

Perhaps a more clear approach (alternative to the above) would be to let \(\alpha = x+iy\) and \(\bar{\alpha} = x-iy\), thus leading to the fact that both
\(a(x+iy)^2+b(x+iy)+c=0\) and \(a(x-iy)^2+b(x-iy)+c=0\) are true. Adding these up, we see that we have \(2a(x^2-y^2)+2bx+c=0\) - given a, x and y are real, b and c must also be real.

Hope this helps :)
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shekhar.patel

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Re: 4U Maths Question Thread
« Reply #2429 on: December 19, 2019, 04:54:51 pm »
0
Hey there!

Could you possibly explain your line of thinking when writing 'since the conjugate roots only applies to imaginary components we can say a, b and c are real?'. I'm a little confused - we could properly correct/understand/explain better if you clarified this :)

Perhaps a more clear approach (alternative to the above) would be to let \(\alpha = x+iy\) and \(\bar{\alpha} = x-iy\), thus leading to the fact that both
\(a(x+iy)^2+b(x+iy)+c=0\) and \(a(x-iy)^2+b(x-iy)+c=0\) are true. Adding these up, we see that we have \(2a(x^2-y^2)+2bx+c=0\) - given a, x and y are real, b and c must also be real.

Hope this helps :)

Hi there, I was trying to just say that the conjugate only affects the imaginary part for example, x= a+ib then x bar = a- ib.
But yes, I see your way of solving the question. Many thanks for the help  :)