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complex numbers watch

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    show tanhz=i has infinitely many solutions in C
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    ln(i) = 0.5ln(-1) = 0.5 i pi
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    For the first one, try thinking if tan(z) is a many-to-one function in C? If it is, then it tan(xz)=i has infinite solutions.
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    sorry that was meant to be tanh(z)

    also, surjective means for all values in domain, there will be a corresponding value in range :p:
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    (Original post by kikzen)
    sorry that was meant to be tanh(z)

    also, injective means one-to-one :p:
    Yeah, I edited it. :p:

    Anyway, same thing applies. If something is many-to-one in C, then it has infinite solutions.
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    (Original post by shift3)
    Yeah, I edited it. :p:

    Anyway, same thing applies. If something is many-to-one in C, then it has infinite solutions.
    ahha so i re-edited mine
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    lol anyway..
    here's a good hint: e^z is many-to-one.
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    tanhz = i

    => [exp(z) - exp(-z)]/[exp(z) + exp(-z)] = i
    => exp(z) - exp(-z) = iexp(z) + iexp(-z)
    => exp(2z) - 1 = iexp(2z) + i
    => exp(2z) - iexp(2z) = 1 + i
    => exp(2z)(1 - i) = 1 + i
    => exp(2z) = (1 + i)/(1 - i)
    => exp(2z) = i

    But i can be written in the form exp(i(pi)/2 + 2k(i)pi), where k is any integer. So,

    exp(2z) = i
    => exp(2z) = exp(i(pi)/2 + 2k(i)pi) (kEZ)
    => 2z = i(pi)/2 + 2k(i)pi
    => z = i(pi)/4 + k(i)pi (kEZ)

    Hence infinitely many solutions in C

    Euclid.
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    gahhh i was writing (1-i)(1+i) as root 2
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    hi again

    draw pictures showing teh sets of complex numbers z such that

    I - The real part of log z is between pi/2 and pi
    II - the imaginary part is between pi/2 and pi (the 'peaks' on this one are where a=b, and its meant to be a curve after that)

    did i get them right?

    argh i just realise the second one is completetly wrong (no idea what i was thinking)
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    you see, i thought if this. that seems too simple a solution

    further to that, if you have z = a+ib and then put it in exp form and log it then the real part is log[rt(a^2 + b^2)] which is how i got my crazy diamond!
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    I)

    a + bi = sqrt(a^2 + b^2)*exp(arctan[b/a]i + 2k(i)pi)
    a + bi = exp(log[sqrt(a^2 + b^2)] + arctan(b/a)i + 2k(i)pi)
    => log(a + bi) = log[sqrt(a^2 + b^2)] + arctan(b/a)i + 2k(i)pi

    Taking the real part only:

    log[sqrt(a^2 + b^2)] <= Pi
    log[sqrt(a^2 + b^2)] >= Pi/2

    => exp(Pi) >= sqrt(a^2 + b^2) >= exp(Pi/2)
    => exp(Pi) >= |z| >= exp(Pi/2) --------------> solutions lie inbetween two circles with radi exp(pi/2) and exp(pi)

    Euclid.
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    ooh that looks a lot nicer than mine at least i was ALONG the right lines, eh!
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    II)

    Equate the imaginary parts to get:

    i) arctan(b/a) <= Pi
    ii) arctan(b/a) >= Pi/2

    This means to say the argument of a and b must lie between Pi and Pi/2, ie all the solutions must lie in the top left quadrant.

    Euclid.
 
 
 
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