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    I have the following example \displaystyle \tan^{-1}\left ( 8 \right ) +\tan^{-1}\left ( 2 \right )+\tan^{-1}\left ( \frac{2}{3} \right ).

    By using arctangent addition, I've gotten to \arctan(0) = n\pi, but the answer is \pi, so how do I eliminate all other solutions?
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    Consider complex numbers say  z_1, z_2, z_3 such that  z_1 z_2 z_3 = w , where  \text{arg}(z_1)= \arctan 8, \text{arg}(z_2)= \arctan 2, \text{arg}(z_3) = \arctan \frac{2}{3} and consider the argument of  w .
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    (Original post by B_9710)
    Consider complex numbers say  z_1, z_2, z_3 such that  z_1 z_2 z_3 = w , where  arg(z_1)= \arctan 8, arg(z_2)= \arctan 2, arg(z_3) = arctan \frac{2}{3} and consider the argument of  w .
    I understand the complex number argument approach, but I'm wonderinh what I'm doing wrong with addition.
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    (Original post by Ayman!)
    I have the following example \displaystyle \tan^{-1}\left ( 8 \right ) +\tan^{-1}\left ( 2 \right )+\tan^{-1}\left ( \frac{2}{3} \right ).

    By using arctangent addition, I've gotten to \arctan(0) = n\pi, but the answer is \pi, so how do I eliminate all other solutions?
    Well, note that for example, \arctan 8, \arctan 2, \arctan \frac{2}{3} are all positive terms, so you know that the sum is bounded below by 0 < n\pi, furthermore, note that each term is bounded above by \frac{\pi}{2} (obviously), so the sum is bounded above by \frac{3\pi}{2} and the only integers that satisfies 0 < n < \frac{3}{2} is n=1.
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    (Original post by Zacken)
    Well, note that for example, \arctan 8, \arctan 2, \arctan \frac{2}{3} are all positive terms, so you know that the sum is bounded below by 0 < n\pi, furthermore, note that each term is bounded above by \frac{\pi}{2} (obviously), so the sum is bounded above by \frac{3\pi}{2} and the only integers that satisfies 0 < n < \frac{3\pi}{2} is n=1.
    That second term of your inequality should be n*pi, not just n, surely?
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    (Original post by HapaxOromenon3)
    That second term of your inequality should be n*pi, not just n, surely?
    Whoop, sorry. Typo there, I meant 0 < n < \frac{3}{2}. Thanks.
 
 
 
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