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Math Help - 1:st year University math

  1. #1
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    1:st year University math

    Hello
    (This may be placed in the wrong part of this forum, so please move it if thats the case.)

    I have just start to read math analysis 1 and i have some problems to understand, so i ask you for some help.

    All solution to:
    (1 + i)z2 − (6 + 4i)z + 9 + 7i = 0,


    All solutions to:
    (z + 1 + 2i)3 = 27i

    If someone can show me how to solve this, i hope i can solve my other problems

    Thanks
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  2. #2
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    Re: 1:st year University math

    Quote Originally Posted by spope View Post
    Hello
    (This may be placed in the wrong part of this forum, so please move it if thats the case.)

    I have just start to read math analysis 1 and i have some problems to understand, so i ask you for some help.

    All solution to:
    (1 + i)z2 − (6 + 4i)z + 9 + 7i = 0,


    All solutions to:
    (z + 1 + 2i)3 = 27i

    If someone can show me how to solve this, i hope i can solve my other problems

    Thanks
    The first is a quadratic, so use the quadratic formula.

    For the second take the cube root of both sides. What is the cube root of i? (Hint: There are three of them. Can you put them into a simple form?)

    -Dan
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  3. #3
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    Re: 1:st year University math

    The first one i just replace z with a+ib and then multiplice?

    And the second is (z+1+2i)=3i?
    Last edited by spope; November 26th 2013 at 11:49 AM.
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  4. #4
    Forum Admin topsquark's Avatar
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    Re: 1:st year University math

    Quote Originally Posted by spope View Post
    The first one i just replace z with a+ib and then multiplice?

    And the second is (z+1+2i)=3i?
    Yeah, it probably would be easier to attack the first problem that way.

    For the second problem, the cube root of i is not i. Rewrite i = e^{i \pi/2 + 2ni \pi}, where n is an integer. What is the cube root of this?

    -Dan
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    Re: 1:st year University math

    Quote Originally Posted by topsquark View Post
    Yeah, it probably would be easier to attack the first problem that way.

    For the second problem, the cube root of i is not i. Rewrite i = e^{i \pi/2 + 2ni \pi}, where n is an integer. What is the cube root of this?

    -Dan
    So (z+1+2i)^{3}= 27 e^{i \pi/2 + 2ni \pi}?

    If it was z^{3}= 27 e^{i \pi/2 + 2ni \pi} i can solve it. But what happen if it is (z+1+2i)^{3}?
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    Re: 1:st year University math

    Hi again

    Thanks for the help topsquark It was pretty easy when i know how to do

    I have some problem solving this one now.


    http://puu.sh/5ZmPT.jpg

    Any idea how to solve from here?
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  7. #7
    Forum Admin topsquark's Avatar
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    Re: 1:st year University math

    Quote Originally Posted by spope View Post
    Hi again

    Thanks for the help topsquark It was pretty easy when i know how to do

    I have some problem solving this one now.


    http://puu.sh/5ZmPT.jpg

    Any idea how to solve from here?
    I presume the exponent "3" in the first line is a typo?

    Let y = cos( \theta ). This turns your equation into a quadratic.

    -Dan
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  8. #8
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    Re: 1:st year University math

    Quote Originally Posted by topsquark View Post
    I presume the exponent "3" in the first line is a typo?

    Let y = cos( \theta ). This turns your equation into a quadratic.

    -Dan
    You are right. It should be sin^2(x)-sqrt(3)cos(x)=7/4


    I have used \sin^2 \theta + \cos^2 \theta = 1 to switch from sin^2(x) to 1-cos^2(x)

    y = cos( \theta )? i don't know how to proceed from there.
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