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  1. #1
    Junior Member Fnus's Avatar
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    Misc Exercises

    1:

    Prove the following:

    zw* - z*w is purely imaginary or zero for all complex numbers z and w

    2:

    A geometric series has a second term of 6 and the sum of its first three terms is -14. Find its fourth term.

    3:

    Is it possible to bend a 20 cm length of wire into the shape of a rectangle which has an area of 30cm2?

    4:

    Find the first two terms of an arithmetic sequence where the sixth term is 21 and the sum of the first seventeen terms is 0.
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  2. #2
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    Quote Originally Posted by Fnus View Post
    1:

    Prove the following:

    zw* - z*w is purely imaginary or zero for all complex numbers z and w
    Let z=x+iy and w=u+iv so \bar z = x - iy and \bar w = u-iv. Thus, z\bar w - \bar z w = ? continue.
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    Quote Originally Posted by Fnus View Post
    1:

    Prove the following:

    zw* - z*w is purely imaginary or zero for all complex numbers z and w
    Hint
    Let  z = x+yi and  w = a + bi
    Then perform the operation zw^* - z^*w , and show it's purely imaginery. Then you can conclude since the choice of imaginery numbers z, w were arbitrary, it is purely imaginery for all z,w.


    A geometric series has a second term of 6 and the sum of its first three terms is -14. Find its fourth term.
    Hint:
    Second term is 6 means you can write ar=6.
    Sum of first 3 is -14 means you can write a+ar+ar^2 = -14
    Solve for a and r and then you can find the 4th term.

    Is it possible to bend a 20 cm length of wire into the shape of a rectangle which has an area of 30cm2?
    Hint:
    Form two simultaneous equations from the information given. One of them is 20 = 2r+2s. Can you find the other? And attempt to solve them, and you'll find that there are no solutions in R.

    4:

    Find the first two terms of an arithmetic sequence where the sixth term is 21 and the sum of the first seventeen terms is 0.
    Hint
    6th term is 21 means you can write 21 = a+5n
    sum of first 17 is 0 means you can write 0 = 17a + 136n

    Solve, and then substitute for first term, a, and second term, a+n.

    __________________________________________________ __________

    Try to solve from the hints. I've written the answers I got below.


































    Answers

    Question 1
     (x+iy)(a-ib) - (x-iy)(a+ib) = 2i(ya+xb) (after simplifying).
    Since y,a,x,b are real, 2i(ya+xb) is imaginery if ya+xb \not = 0

    Question 2
    Solving the simultaneous equations, you'll get r=-3 and a=-2. The 4th term is ar^3 , so  (-2)(-3)^3 = 54 is the 4th term.

    Question 3
    The second equation is 32 = rs
    Solving for s, you'll end up with  s^2 - 10s + 32 =0 . But  b^2 < 4ac \Rightarrow no real solutions, so it's not possible.

    Question 4
    Solving the equations, you'll get a=56, and n=-7.

    So first term, is a which is 56.
    Second term is 56-7 = 49.
    Last edited by WWTL@WHL; February 19th 2008 at 06:45 AM.
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