Z is a complex number.?

I need some help to solve this math question.

Z is a complex number.

Z= a+bi

Z²=x+yi

Prove that 2x²=(a²+b²)^(1/2) +a

7 Answers

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  • Pascal
    Lv 7
    1 decade ago
    Best Answer

    Z²=(a+bi)(a+bi)

    Z²=a²+2abi-b²

    Z²=(a²-b²)+(2ab)i

    x=a²-b²

    2x²=2(a²-b²)²≠(a²+b²)^(1/2) + a (see, for insance, a=1 and b=1)

    So either you wrote the problem wrong or this is supposed to be some kind of trick question, because the identity you're supposed to prove does not actually hold.

  • 1 decade ago

    Square Z:

    a^2+2abi-b^2

    Set it equal to Z^2:

    a^2+2abi-b^2=x+iy

    We know that the real parts and imaginary parts are essentially independent of each other (they satisfy equations independently) so set the real and imaginary parts equal:

    a^2-b^2=x

    2ab=y

    So, from here, I think something in your problem is wrong. The reason I say this is because:

    x^2=(a^2-b^2)^2

    What you have in your problem not only has a plus sign between the two of them (which is possible but not with the other terms you have there) but also is taken the square root of (Which seems really odd). Check your problem and I'll keep an eye on the question.

    I could be wrong, but something about this doesn't feel right.

  • 1 decade ago

    Given:

    Z= a + bi

    Z²= x + yi

    ----------------

    Z= a + bi

    Z²= (a + bi)(a + bi)

    Z²= a² + abi + abi + bi²

    Z²= a² + 2abi + bi²

    But i = √(-1)

    → i² = √(-1)*√(-1)

    → i² = -1

    So:

    Z²= a² + 2abi - b²

    x + yi= a² + 2abi - b² (match up terms).

    → x = a² - b²

    → 2x = 2(a² - b²)

    → yi = 2abi

    → y = 2ab

    This does not match up with what is given in the question.

  • z = a+bi

    z^2 = (a+bi)(a+bi) = aa - bb + 2abi

    z^2 = x+yi

    thus x = aa - bb

    2xx = 2(aa - bb)(aa - bb) = 2a^4 + 2b^4 -4aabb

    but 2xx = (aa+bb)^(1/2)+a, therefore

    2a^4 + 2b^4 -4aabb = (aa+bb)^(1/2)+a

    2a^4 + 2b^4 -4aabb - a = (aa+bb)^(1/2)

    [1] 4(a^4 + b^4 - 2aabb - a)^2 = aa + bb,

    ie, a and b satisfy [1]

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  • Anonymous
    1 decade ago

    Do the obvious multiplication of Z by itself, and collect terms. You have two expressions for Z^2, and by equating them, it should fall right out.

  • 1 decade ago

    a^2-b^2+2abi=x+yi

    x=a^2-b^2

    y=2ab

  • 1 decade ago

    that statement is incorrect.. i tested it on z = 2+3i. please check the textbook or wherever you're getting the question from.

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