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    Hello, I was wondering if anyone could check my solution for this problem, The solution follows as:

    First note that sech^{-1}(i)=\log(\frac{1}{z}+(\frac{1}  {z^2}-1)^\frac{1}{2}). So sech^{-1}(i)=\log(i(-1\pm \sqrt{2}), this means that we have two cases due to the \pm so taking the positive case first we have that sech^{-1}(i)=\log(i(-1+\sqrt{2})), now recall that \log(z)=Log|z|+i(\arg(z)+2k\pi) in this case z=-i+\sqrt{2}i which implies that |z|=-1+\sqrt{2} and \arg(-i+\sqrt{2}i)=\frac{\pi}{2} since there is no real component of z and the imaginary part is greater than 0, so we must have that sech^{-1}(i)=\log(-1+\sqrt{2})+i(\frac{\pi}{2}+2k\p  i) where  k \in \Bbb Z.
    My main concern is the argument i calculated i'm unsure if its correct , i assume if my case is correct for the positive then it will be the same method for the negative. Any help would be highly appreciated.
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    Pricipal argument:
    Yes, your principal argument of  \pi / 2 is fine as  (-1 + \sqrt{2} ) i is purely imaginary and is of form ai, where  a \in \mathbb{R}^{+} .

    As a check, try  \mathrm{sech}(z) , where  z is the answer in your first post.

    Though for part of your last equation, instead of  2ki , you need  2 k \pi .

    Method for negative second case:
    Yes, similar method for the second negative case.

    For the real case, as  \mathrm{sech}(x) is an even function, the positive case is usually used of  \mathrm{arsech}(x) so to return positive values (depending on context of course).
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    (Original post by simon0)
    Pricipal argument:
    Yes, your principal argument of  \pi / 2 is fine as  (-1 + \sqrt{2} ) i is purely imaginary and is of form ai, where  a \in \mathbb{R}^{+} .

    As a check, try  \mathrm{sech}(z) , where  z is the answer in your first post.

    Though for part of your last equation, instead of  2ki , you need  2 k \pi .

    Method for negative second case:
    Yes, similar method for the second negative case.

    For the real case, as  \mathrm{sech}(x) is an even function, the positive case is usually used of  \mathrm{arsech}(x) so to return positive values (depending on context of course).
    Hi thanks for your response I understand most of what you are saying I just don't see how there can be a real case
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    (Original post by simon0)
    Pricipal argument:
    Yes, your principal argument of  \pi / 2 is fine as  (-1 + \sqrt{2} ) i is purely imaginary and is of form ai, where  a \in \mathbb{R}^{+} .

    As a check, try  \mathrm{sech}(z) , where  z is the answer in your first post.

    Though for part of your last equation, instead of  2ki , you need  2 k \pi .

    Method for negative second case:
    Yes, similar method for the second negative case.

    For the real case, as  \mathrm{sech}(x) is an even function, the positive case is usually used of  \mathrm{arsech}(x) so to return positive values (depending on context of course).
    Also what would the Argand diagram look like for the positive case ?
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    (Original post by Scary)
    Hi thanks for your response I understand most of what you are saying I just don't see how there can be a real case
    I meant of  \mathrm{sech}(x) where  x \in \mathbb{R} .
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    (Original post by Scary)
    Also what would the Argand diagram look like for the positive case ?
    Your solutions for the positive case are of form:

     \displaystyle \log ( -1 + \sqrt{2} ) + i( (\pi/2) + 2k \pi ), where  k \in \mathbb{Z} .

    This is just a complex value (of form a + bi) and using the pricipal argument, can you now see how we can create an argand diagram?
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    (Original post by simon0)
    Your solutions for the positive case are of form:

     \displaystyle \log ( -1 + \sqrt{2} ) + i( (\pi/2) + 2k \pi ), where  k \in \mathbb{Z} .

    This is just a complex value (of form a + bi) and using the pricipal argument, can you now see how we can create an argand diagram?
    Yeah the a value is constant so it will just be a set of solutions on that vertical line repeating up and down the imaginary axis
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    (Original post by Scary)
    Yeah the a value is constant so it will just be a set of solutions on that vertical line repeating up and down the imaginary axis
    Yes and of course the set of solutions for the negative case.
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    (Original post by simon0)
    Yes and of course the set of solutions for the negative case.
    Brilliant, thank you for your help, i was just unsure about the diagrams, also a quick question if we had  i^{i\pi} and wanted to find all of it's values in \Bbb C when we plot an argand diagram for these values just all lie on the negative real axis? using the fact that z^{\alpha} =e^{\alpha \log(z)} using this i calculated  i^{i\pi} =e^{-\pi (\frac{\pi}{2}+2k\pi)},k \in \Bbb Z.
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    (Original post by Scary)
    Brilliant, thank you for your help, i was just unsure about the diagrams, also a quick question if we had  i^{i\pi} and wanted to find all of it's values in \Bbb C when we plot an argand diagram for these values just all lie on the negative real axis? using the fact that z^{\alpha} =e^{\alpha \log(z)} using this i calculated  i^{i\pi} =e^{-\pi (\frac{\pi}{2}+2k\pi)},k \in \Bbb Z.
    Looks good to me.
 
 
 
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