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Math Help - Sum of two complex numbers problem

  1. #1
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    Sum of two complex numbers problem

    Hi

    I'm having trouble with the problem below.

    Let z_1 = i and z_2 = \frac{1}{\sqrt{2}} +  \frac{1}{\sqrt{2}}i

    Plot z_1 + z_2 on an Argand diagram and deduce that \tan\left(\frac{3\pi}{8}\right) = 1+\sqrt{2}

    I plotted them and the tangent of the resulting argument was indeed 1+\sqrt{2}, but I have no idea how to show that it is equal to \frac{3\pi}{8}. I've been staring blankly at it for an hour now, I know I'm going to kick myself when I see how it's done! :P

    Stonehambey
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  2. #2
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    Quote Originally Posted by Stonehambey View Post
    Hi

    I'm having trouble with the problem below.

    Let z_1 = i and z_2 = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}i

    Plot z_1 + z_2 on an Argand diagram and deduce that \tan\left(\frac{3\pi}{8}\right) = 1+\sqrt{2}

    I plotted them and the tangent of the resulting argument was indeed 1+\sqrt{2}, but I have no idea how to show that it is equal to \frac{3\pi}{8}. I've been staring blankly at it for an hour now, I know I'm going to kick myself when I see how it's done! :P

    Stonehambey

     z_1+z_2=\frac{1}{\sqrt{2}}+\frac{\sqrt{2}+1}{\sqrt  {2}}i<br />

    Plotting it on the argand diagram . Let the angle of inclination be \theta .Note that this angle has to be expressed in radians .

    Like you , i got \tan\theta=1+\sqrt{2}

    Thus , \theta=tan^{-1}(1+\sqrt{2})

    = 67.5 ( convert it to radian )

     <br />
=67.5\times\frac{\pi}{180}<br />
, giving what you want to find .
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  3. #3
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    How did you get the 67.5? With a calculator? I just assumed there was a more exact way of doing it, unless I'm missing something
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  4. #4
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    Quote Originally Posted by Stonehambey View Post
    How did you get the 67.5? With a calculator? I just assumed there was a more exact way of doing it, unless I'm missing something
    OK ....

    Let O be the point representing z = 0, let A be the point representing z = Re (z_2) = \frac{1}{\sqrt{2}}, let B be the point representing z = z_2 and let C be the point representing z = z_1 + z_2.

    Consider the triangles OAB and OBC. Angle OBA = \frac{\pi}{4} (why?). Therefore angle OBC = \frac{3 \pi}{4}.

    But triangle OBC is isoscles (why?). Therefore angle COB = OCB = \frac{\pi}{8} (why?).

    Therefore angle COA = Arg(z_1 + z_2) = \frac{\pi}{4} + \frac{\pi}{8} = \frac{3 \pi}{8}. Voila!
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