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# complex numbers watch

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1. show tanhz=i has infinitely many solutions in C
2. ln(i) = 0.5ln(-1) = 0.5 i pi
3. 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.
4. sorry that was meant to be tanh(z)

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

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

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

Anyway, same thing applies. If something is many-to-one in C, then it has infinite solutions.
ahha so i re-edited mine
7. lol anyway..
here's a good hint: e^z is many-to-one.
8. 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.
9. gahhh i was writing (1-i)(1+i) as root 2
10. 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)
Attached Images

11. 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!
12. 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.
Attached Images

13. ooh that looks a lot nicer than mine at least i was ALONG the right lines, eh!
14. 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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