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Math Help - z times z

  1. #1
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    z times z

    NOTE: The z in red below has a bar line over it.

    Use the rule z times z = a^2 + b^2 to solve the question below when z = 3 - 4i and w = 8 + 3i.

    Solve z times z =
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by magentarita View Post
    NOTE: The z in red below has a bar line over it.

    Use the rule z times z = a^2 + b^2 to solve the question below when z = 3 - 4i and w = 8 + 3i.

    Solve z times z =
    what are you having trouble with?

    you were given the rule

    here you have a = 3 and b = -4

    just follow the rule
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  3. #3
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    but....

    Quote Originally Posted by Jhevon View Post
    what are you having trouble with?

    you were given the rule

    here you have a = 3 and b = -4

    just follow the rule
    The whole question is: Given z = 3 - 4i, write the question in the form
    a + bi.

    For z times z, the answer is 5(x + 3) (x - 2)^2 (x + 1).

    How do I get this answer?
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by magentarita View Post
    The whole question is: Given z = 3 - 4i, write the question in the form
    a + bi.

    For z times z, the answer is 5(x + 3) (x - 2)^2 (x + 1).

    How do I get this answer?
    ok, i have no idea where you are getting (x + 3) and (x - 2) from. are you leaving out a part of the question?

    the rule is: if z = a + ib

    then

    (1) z^2 = (a + ib)(a + ib) = a^2 - b^2 + 2abi

    (2) |z|^2 = a^2 + b^2

    where are your x's coming from?

    EDIT: Oh! i see now, you put red z to mean \bar{z}!

    ok, so z \bar{z} = (a + ib)(a - ib) = a^2 + b^2 = |z|^2

    still, where are you getting your x's from?!
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  5. #5
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    Quote Originally Posted by magentarita View Post
    The whole question is: Given z = 3 - 4i, write the question in the form
    a + bi.

    For z times z, the answer is 5(x + 3) (x - 2)^2 (x + 1).

    How do I get this answer?
    What you need to do is go back and read the question CAREFULLY.

    For one thing it surely does not say "write the QUESTION in the form a+bi". Perhaps it says "write the ANSWER in the form a+ bi.

    But then "5(x+ 3)(x-2)^2(x+1)" is NOT in the form a+ bi!

    As you said before z\overline{z}= 3^2+ (-4)^2, a positive real number, not some product of algebraic expressions. Perhaps you accidently looked up the answer to the wrong question.
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  6. #6
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    the x letters

    Quote Originally Posted by Jhevon View Post
    ok, i have no idea where you are getting (x + 3) and (x - 2) from. are you leaving out a part of the question?

    the rule is: if z = a + ib

    then

    (1) z^2 = (a + ib)(a + ib) = a^2 - b^2 + 2abi

    (2) |z|^2 = a^2 + b^2

    where are your x's coming from?

    EDIT: Oh! i see now, you put red z to mean \bar{z}!

    ok, so z \bar{z} = (a + ib)(a - ib) = a^2 + b^2 = |z|^2

    still, where are you getting your x's from?!
    I think the x letters could be a typo in the math book. Math books are famous for having tons of wrong answers to great questions.
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  7. #7
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by magentarita View Post
    I think the x letters could be a typo in the math book. Math books are famous for having tons of wrong answers to great questions.
    not all math books, but some do. this is definitely a typo (are you sure you're looking in the right section?) there is no where for an x to appear.
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  8. #8
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    it was a..............

    Quote Originally Posted by HallsofIvy View Post
    What you need to do is go back and read the question CAREFULLY.

    For one thing it surely does not say "write the QUESTION in the form a+bi". Perhaps it says "write the ANSWER in the form a+ bi.

    But then "5(x+ 3)(x-2)^2(x+1)" is NOT in the form a+ bi!

    As you said before z\overline{z}= 3^2+ (-4)^2, a positive real number, not some product of algebraic expressions. Perhaps you accidently looked up the answer to the wrong question.
    It was a textbook error.
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  9. #9
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    great work.......

    Quote Originally Posted by Jhevon View Post
    what are you having trouble with?

    you were given the rule

    here you have a = 3 and b = -4

    just follow the rule
    Fantastic!
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  10. #10
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    I see..............

    Quote Originally Posted by Jhevon View Post
    ok, i have no idea where you are getting (x + 3) and (x - 2) from. are you leaving out a part of the question?

    the rule is: if z = a + ib

    then

    (1) z^2 = (a + ib)(a + ib) = a^2 - b^2 + 2abi

    (2) |z|^2 = a^2 + b^2

    where are your x's coming from?

    EDIT: Oh! i see now, you put red z to mean \bar{z}!

    ok, so z \bar{z} = (a + ib)(a - ib) = a^2 + b^2 = |z|^2

    still, where are you getting your x's from?!
    I got it now. Thank you.
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  11. #11
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    yes.........

    Quote Originally Posted by Jhevon View Post
    not all math books, but some do. this is definitely a typo (are you sure you're looking in the right section?) there is no where for an x to appear.
    Yes, I looked at the section several times before deciding to place the question in this forum.
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