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Thread: Complex arctan

  1. #1
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    Complex arctan

    I found these two questions in a textbook and they are driving me nuts.

    How can you derive from $\displaystyle (1+i)(2+i)(3+i)=10i$ that $\displaystyle \arctan1+\arctan\frac{1}{2}+\arctan\frac{1}{3}=\fr ac{\pi}{2}$?

    Similarly, how can you with the help of $\displaystyle (1-2i)^4 = -7+24i$ show that $\displaystyle 4\arctan2=\pi+\arctan\frac{24}{7}$
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  2. #2
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    Quote Originally Posted by Mondreus View Post
    I found these two questions in a textbook and they are driving me nuts.

    How can you derive from $\displaystyle (1+i)(2+i)(3+i)=10i$ that $\displaystyle \arctan1+\arctan\frac{1}{2}+\arctan\frac{1}{3}=\fr ac{\pi}{2}$?
    1+ i can be written, in polar form, as $\displaystyle \sqrt{2}(cos(\theta_1)+ i sin(\theta_1)$ where $\displaystyle tan(\theta_1)= 1/1= 1$ (and, of course, $\displaystyle \theta_1= \pi/4$). 2+ i can be written as $\displaystyle \sqrt{5}(cos(\theta_2)+ i sin(\theta_2))$ where $\displaystyle tan(\theta_2)= 1/2$. 3+ i can be written as $\displaystyle \sqrt{10}(cos(\theta_3)+ i sin(\theta_3)$ where $\displaystyle tan(\theta_3)= 1/3$.

    By DeMoivre's formula, their product is $\displaystyle (\sqrt{2})(\sqrt{5})\sqrt{10}(cos(\theta_1+ \theta_2+ \theta_3)+ i sin(\theta_ 3+ \theta_2+ \theta_3))$.

    The fact that the product is 10 i tells you that $\displaystyle cos(\theta_1+ \theta_2+ \theta_3)= 0$ and $\displaystyle cos(\theta_1+ \theta_2+ \theta_3)= 1$ or that $\displaystyle \theta_1+ \theta_2+ \theta_3= \pi/2$.

    Similarly, how can you with the help of $\displaystyle (1-2i)^4 = -7+24i$ show that $\displaystyle 4\arctan2=\pi+\arctan\frac{24}{7}$
    $\displaystyle 1- 2i= \sqrt{5}(cos(\theta+ i sin(\theta)$ where $\displaystyle tan(\theta_1)= 2$ and, again by DeMoivre's formula, that is equal to $\displaystyle 25(cos(4\theta)+ i sin(4\theta))$. -7+ 24i can be written as $\displaystyle 25 (cos(\phi)+ i sin(\phi))$ where $\displaystyle tan(\phi)= 24/7$.
    Last edited by HallsofIvy; Dec 26th 2009 at 06:08 AM.
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    What about the $\displaystyle \pi$ part in the last problem?
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    Consider which quadrant you are in.
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  5. #5
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    Quote Originally Posted by HallsofIvy View Post
    1+ i can be written, in polar form, as $\displaystyle \sqrt{2}(cos(\theta_1)+ i sin(\theta_1)$ where $\displaystyle tan(\theta_1)= 1/1= 1$ (and, of course, $\displaystyle \theta_1= \pi/4$). 2+ i can be written as $\displaystyle \sqrt{5}(cos(\theta_2)+ i sin(\theta_2))$ where $\displaystyle tan(\theta_2)= 1/2$. 3+ i can be written as $\displaystyle \sqrt{10}(cos(\theta_3)+ i sin(\theta_3)$ where $\displaystyle tan(\theta_3)= 1/3$.

    By DeMoivre's formula, their product is $\displaystyle (\sqrt{2})(\sqrt{5})\sqrt{10}(cos(\theta_1+ \theta_2+ \theta_3)+ i sin(\theta_ 3+ \theta_2+ \theta_3))$.

    The fact that the product is 10 i tells you that $\displaystyle cos(\theta_1+ \theta_2+ \theta_3)= 0$ and $\displaystyle cos(\theta_1+ \theta_2+ \theta_3)= 1$ or that $\displaystyle \theta_1+ \theta_2+ \theta_3= \pi/2$.


    $\displaystyle 1- 2i= \sqrt{5}(cos(\theta+ i sin(\theta)$ where $\displaystyle tan(\theta_1)= 2$ and, again by DeMoivre's formula, that is equal to $\displaystyle 25(cos(4\theta)+ i sin(4\theta))$. -7+ 24i can be written as $\displaystyle 25 (cos(\phi)+ i sin(\phi))$ where $\displaystyle tan(\phi)= 24/7$.

    Wow this problem is confusing. It took me a little bit to figure it out.

    HallsofIvy you have a typo when you said


    $\displaystyle cos(\theta_1+ \theta_2+ \theta_3)= 1$

    it should be

    $\displaystyle sin(\theta_1+ \theta_2+ \theta_3)= 1$

    Thanks for working it out though. It really did help me figure out my mistake.


    I kept working out

    $\displaystyle 4\arctan(-2)=\pi+\arctan\bigg(-\frac{24}{7}\bigg)$

    Since the original problem was in the second quad. Didn't notice the tan had been changed to positive value changing the problem.
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    Quote Originally Posted by 11rdc11 View Post
    Wow this problem is confusing. It took me a little bit to figure it out.

    HallsofIvy you have a typo when you said


    $\displaystyle cos(\theta_1+ \theta_2+ \theta_3)= 1$

    it should be

    $\displaystyle sin(\theta_1+ \theta_2+ \theta_3)= 1$
    Yes, thanks for the correction.

    Thanks for working it out though. It really did help me figure out my mistake.


    I kept working out

    $\displaystyle 4\arctan(-2)=\pi+\arctan\bigg(-\frac{24}{7}\bigg)$

    Since the original problem was in the second quad. Didn't notice the tan had been changed to positive value changing the problem.
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