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Math Help - Complex No.

  1. #1
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    Complex No.

    hi...

    appreciate if anyone can help me with this complex no. question....especially bi)....need detailed explanation!

    Ans as follows:

    (b) (i) 2.21
    (ii) -2 + I , -2 -i

    Thank You!
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by shaojie2k View Post
    hi...

    appreciate if anyone can help me with this complex no. question....especially bi)....need detailed explanation!

    Ans as follows:

    (b) (i) 2.21
    (ii) -2 + I , -2 -i

    Thank You!
    b) The solution set for |z+2|=1 is a circle of radius 1 centered on the real number z = -2 in the Argand plane. There is no single solution. The representative solution would be:
    z = (cos \theta - 2) + i \, sin \theta

    -Dan
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by shaojie2k View Post
    hi...

    appreciate if anyone can help me with this complex no. question....especially bi)....need detailed explanation!

    Ans as follows:

    (b) (i) 2.21
    (ii) -2 + I , -2 -i

    Thank You!
    a)
    \frac{w^2+4}{4-w^2} = i

    w^2 + 4 = 4i - iw^2

    (1+i)w^2 = -4 + 4i

    w^2 = 4 \frac{-1+i}{1+i}

    The simplest way to proceed from here would be to transform the RHS to the form re^{i \theta}.

    \frac{-1+i}{1+i} = \frac{-1+i}{1+i} \cdot \frac{1-i}{1-i} = \frac{-(1-i)^2}{1+1} = -\frac{1-2i-1}{2} = i = e^{\pi \, i/2}

    So
    w^2 = e^{\pi \, i/2}

    w = \pm e^{\pi \, i/4}

    So one solution for w is e^{\pi \, i/4} and the other is -e^{\pi \, i/4} = e^{\pi \, i} \cdot e^{\pi \, i/4} = e^{5 \pi \, i/4} = e^{-2 \pi \, i} \cdot e^{5 \pi \, i/4} = e^{-3 \pi \, i/4}. (The last trick with the e^{-2 \pi \, i} is simply to force -\pi < Arg(w) < \pi.)
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by shaojie2k View Post
    hi...

    appreciate if anyone can help me with this complex no. question....especially bi)....need detailed explanation!

    Ans as follows:

    (b) (i) 2.21
    (ii) -2 + I , -2 -i

    Thank You!
    i)
    I'm having a problem with this. My approach is to write z + i = re^{i \theta} + i = Re^{i \Theta}

    I got that \Theta = atn \left ( \frac{1+sin \theta}{cos \theta} \right )

    The problem is that this function has no minimum:

    \frac{d \Theta}{d \theta} = \frac{1}{\left ( \frac{1+sin \theta}{cos \theta} \right )^2 + 1} \cdot \left [ 1 + \frac{sin^2 \theta + sin \theta}{cos^2 \theta} \right ] = \frac{1 + sin \theta}{2(sin \theta + 1)} = \frac{1}{2}

    which can never be zero.

    Now, looking at the graph of the function \Theta( \theta ) you might think that there is a local minimum at \theta = \pi /2, but the function does not exist there.

    (I'm going to attempt to upload the graph. Cross your fingers!)

    (Yay! It worked! )

    -Dan
    Attached Thumbnails Attached Thumbnails Complex No.-theta1.jpg  
    Last edited by topsquark; October 22nd 2006 at 08:02 AM. Reason: typo
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  5. #5
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by shaojie2k View Post
    hi...

    appreciate if anyone can help me with this complex no. question....especially bi)....need detailed explanation!

    Ans as follows:

    (b) (i) 2.21
    (ii) -2 + I , -2 -i

    Thank You!
    ii) Again, I'm having a problem.

    Write z = re^{i \, \theta} = r \, cos \theta + ir \, sin \theta

    Then z - z^* = 2ir \, sin \theta

    |z - z^*| = 2r |sin \theta |

    This has no maximum since we may choose r to be as large as we wish.

    -Dan
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