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Math Help - Complex numbers: 2 problems.

  1. #1
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    Complex numbers: 2 problems.

    Hi, I have two problems dealing with complex numbers, ill show you them and then what I try to do:

    Given z = x + iy, i = \sqrt-1 So yeah y = the imaginary and z = x(real)

    1. z + 1 - i = (1+6) / (3 - i)  Answer given: x = -4/5, y =7/5

    ^ This one I multiplied by the conjugate 3 + i on the right side. I got:

    z + 1 - i = (21 + 7i)/ 10 and now do I'm unsure what to do.

    2. (2 + \sqrt-2)(1 - \sqrt-8)

    I'm not sure if I know how to multiply this out correctly...

    2 - 2\sqrt-8 - \sqrt-2 + 2\sqrt4?
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  2. #2
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    Quote Originally Posted by Kaln0s View Post
    Hi, I have two problems dealing with complex numbers, ill show you them and then what I try to do:

    Given z = x + iy, i = \sqrt-1 So yeah y = the imaginary and z = x(real)

    1. z + 1 - i = (1+6) / (3 - i)  Answer given: x = -4/5, y =7/5

    ^ This one I multiplied by the conjugate 3 + i on the right side. I got:

    z + 1 - i = (21 + 7i)/ 10 and now do I'm unsure what to do.

    2. (2 + \sqrt-2)(1 - \sqrt-8)

    I'm not sure if I know how to multiply this out correctly...

    2 - 2\sqrt-8 - \sqrt-2 + 2\sqrt4?
    For problem 1. it might have been easier to multiply both sides of the equation by (3-i), as opposed to multiplying the RHS by (3+i)/(3+i). I suggest you try that strategy!

    2. Just use foil, but remember that  \sqrt{-a} = i\sqrt{a} , hence your expression should be:

    (2 + i\sqrt{2})(1 - i\sqrt{8}) = 2 -2i\sqrt{8}+i\sqrt{2}-i^2 \sqrt{2}.\sqrt{8}

    Then remember that  i^2 = \sqrt{-1}^2 = -1 and that  \sqrt{a}.\sqrt{b} = \sqrt{ab} , hence:

    = 2 -2i\sqrt{8}+i\sqrt{2}+ \sqrt{16}

    = 2 -2i\sqrt{8}+i\sqrt{2}+ 4

    Now split into real and imaginary parts!

    = (2+4) +i(\sqrt{2}-2\sqrt{8})

    PS: Are you sure you wrote out 1. correctly? Since the numerator of the RHS is just 6+1 = 7.
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  3. #3
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    Quote Originally Posted by Kaln0s View Post
    Hi, I have two problems dealing with complex numbers, ill show you them and then what I try to do:

    Given z = x + iy, i = \sqrt-1 So yeah y = the imaginary and z = x(real)

    1. z + 1 - i = (1+6) / (3 - i)  Answer given: x = -4/5, y =7/5

    ^ This one I multiplied by the conjugate 3 + i on the right side. I got:

    z + 1 - i = (21 + 7i)/ 10 and now do I'm unsure what to do.

    2. (2 + \sqrt-2)(1 - \sqrt-8)

    I'm not sure if I know how to multiply this out correctly...

    2 - 2\sqrt-8 - \sqrt-2 + 2\sqrt4?

    1. recheck the problem ... is the (1+6) in the numerator on the right side correct?


    2. (2 + i\sqrt{2})(1 - 2i\sqrt{2}) =

    2 + i\sqrt{2} - 4i\sqrt{2} + 4 =

    6 - 3i\sqrt{2}
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  4. #4
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    Quote Originally Posted by skeeter View Post
    1. recheck the problem ... is the (1+6) in the numerator on the right side correct?


    2. (2 + i\sqrt{2})(1 - 2i\sqrt{2}) =

    2 + i\sqrt{2} - 4i\sqrt{2} + 4 =

    6 - 3i\sqrt{2}
    I uploaded both problems, and they both show the answer. The deal is it's a really bad printout, so you tell me if it's 1+6 :P.
    Attached Thumbnails Attached Thumbnails Complex numbers: 2 problems.-36.jpg   Complex numbers: 2 problems.-38.jpg  
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