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    Could someone please help with these questions?

    Find the fourth roots of 1/2 +1/2i

    Verify that -i is a root of x^4-5x^3+7x^2-5x+6

    Write in the form x+iy:
    i^2015 is that just 1?

    e^(ipi/4)
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    Hello, I've moved this to maths study help for you.


    (Original post by elef95)

    Find the fourth roots of 1/2 +1/2i
    Do you know the method to find the roots?

    (Original post by elef95)
    Verify that -i is a root of x^4-5x^3+7x^2-5x+6
    Have you got any thoughts on how to do this?
    (Original post by elef95)
    i^2015 is that just 1?
    It's not 1. Do you know the trick where powers of i repeat in a cycle of 4? i^0=1, i^1=i, i^2=-1, i^3=-i, i^4=1,.... If you do 2015/4=503.75 and just focus on the stuff after the decimal point, you get 0.75. If you do 0.75*4 you get 3, so this tells us that i^{2015}=i^3=-i. (I can do more examples of this if it's not clear)

    (Original post by elef95)
    e^(ipi/4)
    Have you made a start on this one?
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    (Original post by rayquaza17)
    Hello, I've moved this to maths study help for you.

    Do you know the method to find the roots?
    Do you write it in polar form
    z= sqrt2/2(cos45+isin45)
    and then
    z=(sqrt2/2)^1/4(cos45/4+isin45/4)
    (Original post by rayquaza17)
    Have you got any thoughts on how to do this?
    I'm not sure how to go about answering this one.
    (Original post by rayquaza17)
    It's not 1. Do you know the trick where powers of i repeat in a cycle of 4? i^0=1, i^1=i, i^2=-1, i^3=-i, i^4=1,.... If you do 2015/4=503.75 and just focus on the stuff after the decimal point, you get 0.75. If you do 0.75*4 you get 3, so this tells us that i^{2015}=i^3=-i. (I can do more examples of this if it's not clear)
    I think I understand this now. Thank you.
    (Original post by rayquaza17)
    Have you made a start on this one?
    Do you use Euler's equation?
    e^ipi/4 = cos pi/4 + isin pi/4
    Do you take the natural log of both sides to get rid of the e?
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    (Original post by elef95)
    I'm not sure how to go about answering this one.
    If you want to verify that z=a is a root of a polynomial p(z), you need only show that p(a) = 0.
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     i^{2015}= (i^4)^{503} \times i^3 .
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    Sorry it took me so long to reply, I was at work.
    (Original post by elef95)
    Do you write it in polar form
    z= sqrt2/2(cos45+isin45)
    and then
    z=(sqrt2/2)^1/4(cos45/4+isin45/4)
    This is very close, but you need to remember that when we take the nth root of a complex number, there are exactly n distinct roots. So here, we expect 4 different numbers.

    The general formula is on the second slide of here: http://www-thphys.physics.ox.ac.uk/p...x_11_4_cls.pdf

    (sorry I tried to write it in latex but it keeps messing up, and my dinner is ready lol)

    So you basically want to find r, find theta (in radians) and then sub it into there and change the value of k to find the different roots.

    (Original post by elef95)
    Do you use Euler's equation?
    e^ipi/4 = cos pi/4 + isin pi/4
    Do you take the natural log of both sides to get rid of the e?
    Close, all you need to do from here is work out in your calculator (or by memory!) what cos(pi/4) is and what sin(pi/4) is and tells you what z=x+iy are. (Because x=cos(pi/4), y=sin(pi/4) in this case).
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    (Original post by Zacken)
    If you want to verify that z=a is a root of a polynomial p(z), you need only show that p(a) = 0.
    I don't understand
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    (Original post by elef95)
    I don't understand
    If you want to show that p(z) = z^3 - z let's say has a root at z=1 then you just need to show that p(1) = 1^3 - 1 = 0. Hence z=1 is a root. Apply this to your function with x=-i and show that if you plug that into your polynomial, you get 0 out.
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    (Original post by Zacken)
    If you want to show that p(z) = z^3 - z let's say has a root at z=1 then you just need to show that p(1) = 1^3 - 1 = 0. Hence z=1 is a root. Apply this to your function with x=-i and show that if you plug that into your polynomial, you get 0 out.
    Oh ok thank you I get it!
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    (Original post by elef95)
    Oh ok thank you I get it!
    No worries, what about Rayquaza's post? Have you understood all your questions now?
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    (Original post by Zacken)
    No worries, what about Rayquaza's post? Have you understood all your questions now?
    For the fourth roots of 1/2+1/2i question what do you change what you get the cos and sin of by for each root?
    I attached what I did.
    Attached Images
     
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    (Original post by rayquaza17)
    This is very close, but you need to remember that when we take the nth root of a complex number, there are exactly n distinct roots. So here, we expect 4 different numbers.
    I don't know what I'm doing wrong. I attached what I did for this question in the above post. Can you see what I did wrong?
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    (Original post by elef95)
    I don't know what I'm doing wrong. I attached what I did for this question in the above post. Can you see what I did wrong?
    Sorry I'll have a look tonight when I get home
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    (Original post by elef95)
    For the fourth roots of 1/2+1/2i question what do you change what you get the cos and sin of by for each root?
    I attached what I did.
    In the coses and sines you have (45/4)*(1/4) so you are dividing by 4 twice - you only need 45/4, not the (1/4) as well.

    You also might need to change the 45 degrees into radians.
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    (Original post by rayquaza17)
    In the coses and sines you have (45/4)*(1/4) so you are dividing by 4 twice - you only need 45/4, not the (1/4) as well.

    You also might need to change the 45 degrees into radians.
    Thank you!
 
 
 
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