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Math Help - [SOLVED] easy complex number questions

  1. #1
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    Exclamation [SOLVED] easy complex number questions

    hi, could someone check the answers of these for me? btw, the book says that they are all wrong. i dont understand why.

    1. If z = 2 + i, evaluate:

    a) 3z = 6 + 3i
    b)  (z-1)(z^2 + z +1)
     = z^3 - 1
    = (2+i)^3
    = (2 + i)^2(2+i)
    = (4 + i^2 + 4i)(2+2i)
    = (3 + 4i) (2+2i)
    = 6 + 8i + 6i + 8i^2
    = -2 + 14i

    2. Solve for z; expressing answers in the form a + ib
    a)  \frac{z+3}{z - 1} = 2 - 3i
     \frac{z -1 + 4}{z-1} = 2-3i
     1 + \frac{4}{z-1} = 2 - 3i
     2 - 3i -1 = \frac{4}{z-1}
    1 - 3i = \frac{4}{z -1}
    \frac{4}{1-3i} + 1 = z
     \frac{4(1 + 3i)}{(1-3i)(1+3i)}+ 1 =z
     \frac{4 + 12i}{10} + 1 = z
    \frac{4+12i + 10}{10} = z
     \frac{14 + 12i}{10} = z
     \frac{7}{5} + \frac{6}{5}i = z

    b) Totally clueless about this one:
    Solve for x.  \frac{2z}{2+i} + 3- 2i = (1-i)z

    3. Simplify:  \frac{\sqrt{5+12i}+ \sqrt{5-12i}}{\sqrt{5+12i}-\sqrt{5-12i}}

    I got 3i/2. but the answer is -3i/2. I dont get why.

    Thanks
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  2. #2
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    Exclamation [SOLVED] solving complex number equation

    Solve for z. \frac{2z}{2+i} + 3- 2i = (1-i)z

    I tried it but nothing gets rid of the z.. could someone give me a hint? I multiplied by 2 + i on both sides, but that didn't work. did a whole lot of algebra bashing, but nothing gets rid of the 2z on the left hand side.

    Quote Originally Posted by mathaddict
    HI

    Let z = a + ib

    \frac{2(a+bi)}{2+i}+3-2i=(1-i)(a+bi)

    Multiply by its conjugate,

    <br />
\frac{(2a+2bi)(2-i)}{5}+3-2i=(a+b)+(b-a)i<br />

    4a+2b+15-2ai+4bi-10i=5(a+b)+5(b-a)i

    By comparing ,

    4a+2b+15=5(a+b).... 1

    4b-2a-10=5(b-a).... 2

    Now solve the system for a and b
    ..
    Last edited by mr fantastic; October 24th 2009 at 02:38 PM. Reason: Question and reply moved from another thread. And moved back. Sorry for the confusion.
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  3. #3
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    Quote Originally Posted by differentiate View Post

    b) Totally clueless about this one:
    Solve for x.  \frac{2z}{2+i} + 3- 2i = (1-i)z

    3. Simplify:  \frac{\sqrt{5+12i}+ \sqrt{5-12i}}{\sqrt{5+12i}-\sqrt{5-12i}}

    I got 3i/2. but the answer is -3i/2. I dont get why.

    Thanks
    your sum 1(a), 1(b),2(a) are right
    but 2(b) solve for x
     \frac{2z}{2+i} + 3- 2i = (1-i)z
    where is x (?) in the statement ?
    for (3) the answer is -3i/2 .please show your work so that we can find mistakes?
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  4. #4
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    Help me find the mistake in my working out. I got 3i/2, however the answer is -3i/2.

    Evaluate: \frac{\sqrt{5+12i} + {\sqrt{5-2i}}}{\sqrt{5+12i} - {\sqrt{5-12i}}}

     \sqrt{5+12i} = a + bi
     5 + 12i = a^2 + b^2i^2 + 2abi
     5 + 12i = a^2 - b^2 + 2abi
    2ab = 12, a^2 - b^2 = 5
     b = 6/a
     a^2 - \frac{6^2}{a^2} = 5
    a^4 - 5a^2 - 36 = 0
     (a^2 - 9)(a^2 +4) = 0
    a^2 = \pm3
     \sqrt{5+12i} = \pm (3+12i)

     \sqrt{5 - 12i} = a + bi
    5 - 12i = a^2 + b^2i^2 + 2abi
     5 - 12i = a^2 - b^2 + 2abi
    2ab = -12, a^2 - b^2 = 5
     b = \frac{-6}{a}
     a^2 - (\frac{-6}{a})^2 = 5
    a^4 - 5a^2 - 36 = 0
     (a^2 - 9)(a^2 + 4)=0
     a = \pm 3
     \sqrt{5 - 12i} = \pm (3-2i)

    hence, \frac{\sqrt{5+12i} + {\sqrt{5-2i}}}{\sqrt{5+12i} - {\sqrt{5-12i}}}
      = \frac{3 + 2i + 3 - 2i}{3 + 2i - 3 + 2i} = \frac{6}{4i} = \frac{3}{2i}
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  5. #5
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    You have

    \frac{\sqrt{5+12i}+\sqrt{5-12i}}{\sqrt{5+12i}-\sqrt{5-12i}} = \frac{\sqrt{5+12i}+\sqrt{5-12i}}{\sqrt{5+12i}-\sqrt{5-12i}}\cdot \frac{\sqrt{5+12i}+\sqrt{5-12i}}{\sqrt{5+12i}+\sqrt{5-12i}}

    This is just

    \frac{5+12i + 2 \sqrt{5-12i}\sqrt{5+12i}+5-12i}{5+12i-5+12i}

    Try this -- you might find it easier.
    Last edited by harbottle; October 23rd 2009 at 10:55 PM.
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  6. #6
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    Quote Originally Posted by harbottle View Post
    You have

    \frac{\sqrt{5+12i}+\sqrt{5-12i}}{\sqrt{5+12i}-\sqrt{5-12i}} = \frac{\sqrt{5+12i}+\sqrt{5-12i}}{\sqrt{5+12i}-\sqrt{5-12i}}\cdot \frac{\sqrt{5+12i}+\sqrt{5-12i}}{\sqrt{5+12i}+\sqrt{5-12i}}

    This is just

    \frac{5+12i + 2 \sqrt{5-12i}\sqrt{5+12i}+5-12i}{5+12i-5+12i}

    Try this -- you might find it easier.
    Then to perform the complex division so you don't have a complex denominator, multiply top and bottom by the bottom's conjugate.
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  7. #7
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    alright so then  \frac{10 + 2\sqrt{169}}{24i}
     \frac{36}{24i}
    \frac{36i}{24i^2}
    \frac{-3}{2}
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  8. #8
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    Quote Originally Posted by differentiate View Post
    alright so then  \frac{10 + 2\sqrt{169}}{24i}
     \frac{36}{24i}
    \frac{36i}{24i^2}
    \frac{-3}{2}

    okay, so my question is... the method I used before.. like by finding the square roots and stuff. Why didn't that work?
    Your working out is correct here...

    I don't understand your question...
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  9. #9
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    alright, thanks for the help. I finally got the answer.
    Last edited by mr fantastic; October 24th 2009 at 05:05 AM. Reason: Moved a question to a new thread.
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