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Thread: 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 $\displaystyle |z+2|=1$ 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:
    $\displaystyle z = (cos \theta - 2) + i \, sin \theta$

    -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)
    $\displaystyle \frac{w^2+4}{4-w^2} = i$

    $\displaystyle w^2 + 4 = 4i - iw^2$

    $\displaystyle (1+i)w^2 = -4 + 4i$

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

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

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

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

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

    So one solution for w is $\displaystyle e^{\pi \, i/4}$ and the other is $\displaystyle -e^{\pi \, i/4} = e^{\pi \, i} \cdot e^{\pi \, i/4} = e^{5 \pi \, i/4} = e^{-2 \pi \, i} \cdot e^{5 \pi \, i/4} = e^{-3 \pi \, i/4}$. (The last trick with the $\displaystyle e^{-2 \pi \, i}$ is simply to force $\displaystyle -\pi < Arg(w) < \pi$.)
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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 $\displaystyle z + i = re^{i \theta} + i = Re^{i \Theta}$

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

    The problem is that this function has no minimum:

    $\displaystyle \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 ] $ = $\displaystyle \frac{1 + sin \theta}{2(sin \theta + 1)} = \frac{1}{2}$

    which can never be zero.

    Now, looking at the graph of the function $\displaystyle \Theta( \theta )$ you might think that there is a local minimum at $\displaystyle \theta = \pi /2$, 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; Oct 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 $\displaystyle z = re^{i \, \theta} = r \, cos \theta + ir \, sin \theta$

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

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

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

    -Dan
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