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Inverse trig functions + complex numbers watch

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    My (Heinemann FP3 for Edexcel) textbook asks in an exercise to "find arccos 2, in the form a + ib." By letting z = arccos 2, and then cossing both sides and using the complex exponential definition of cosine and then the quadratic formula, I found that

    e^{iz} = 4 \pm \sqrt{15}
    Now by taking logs normally (and noting that the positive root is the reciprocal of the negative root), I get
    iz = \pm \ln(4 + \sqrt{15})
    which gives the answer
    \mathrm{arccos} 2 = \pm i \ln(4 + \sqrt{15}).

    However, the textbook gives the answer as
    2n \pi \pm i \ln(4 + \sqrt{15}).

    This raises a few questions:
    - How are the inverse trig functions defined for arguments outside of the usual domain of the function? Are they well defined at all?
    - Why is the function multi-valued? I would expect it to take some sort of "principal value", similar to your normal arccos/arcsin functions.
    - How did the textbook expect me to get to its given answer (including knowing that it seems to have a period of 2 \pi even for complex arguments) without giving me any examples or methods for doing this sort of question, or defining the function I'm using for the argument it's given me? My answer should really be: "arccos 2 is undefined, because in book C3 the domain for the arccos function is given as -1 \leq x \leq 1."
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    - You can the other solutions by remembering that e^{iz} = e^{iz + 2n\pi} for all n \in \mathbb{Z}

    - I'm not sure what the Principal value for complex trig functions, but that I've never come across it suggests it probably isn't that important, or used a great deal.

    - In this question, we need not really worry if it is a function or not. All we are asked to do is find the values such that \cos z = 2. Usually we like to stick within our safe functions, which we know lots about, but other times we are just trying to find what numbers satisfy some property.
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    (Original post by tommm)
    My (Heinemann FP3 for Edexcel) textbook asks in an exercise to "find arccos 2, in the form a + ib." By letting z = arccos 2, and then cossing both sides and using the complex exponential definition of cosine and then the quadratic formula, I found that

    e^{iz} = 4 \pm \sqrt{15}
    Now by taking logs normally (and noting that the positive root is the reciprocal of the negative root), I get
    iz = \pm \ln(4 + \sqrt{15})
    which gives the answer
    \mathrm{arccos} 2 = \pm i \ln(4 + \sqrt{15}).

    However, the textbook gives the answer as
    2n \pi \pm i \ln(4 + \sqrt{15}).

    This raises a few questions:
    - How are the inverse trig functions defined for arguments outside of the usual domain of the function? Are they well defined at all?
    - Why is the function multi-valued? I would expect it to take some sort of "principal value", similar to your normal arccos/arcsin functions.
    - How did the textbook expect me to get to its given answer (including knowing that it seems to have a period of 2 \pi even for complex arguments) without giving me any examples or methods for doing this sort of question, or defining the function I'm using for the argument it's given me? My answer should really be: "arccos 2 is undefined, because in book C3 the domain for the arccos function is given as -1 \leq x \leq 1."
    \cos x is many-to-one, and \arccos y is one-to-one but you in some cases would need to find a solution which would result in "one-to-many", if it is not exactly specified. For example \cos \pi = -1, but \cos {-\pi} = -1 also. So we do y = 2n \pi \pm \pi to find the one-to-many.
    You're supposed to think to figure this out, as you might not learn as much otherwise :p:.
    A C3 book may also say that your arccos function is in the real x range -1 to 1, whereas complex numbers expand beyond these assumptions in FP3 or so .
 
 
 
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