The Student Room Group
Uni students - Complete our survey >>
Uni students - Complete our survey >>
Uni students - Complete our survey >>

This discussion is now closed.

Check out other Related discussions

complex numbers

show tanhz=i has infinitely many solutions in C
Reply 1
Avatar for shift3
shift3
ln(i) = 0.5ln(-1) = 0.5 i pi
Reply 2
Avatar for shift3
shift3
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.
Reply 3
Avatar for kikzen
kikzen
OP
14
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:
Reply 4
Avatar for shift3
shift3
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.
Reply 5
Avatar for kikzen
kikzen
OP
14
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
Reply 6
Avatar for shift3
shift3
lol anyway..
here's a good hint: e^z is many-to-one.
Reply 7
Avatar for Gauss
Gauss
11
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.
Reply 8
Avatar for kikzen
kikzen
OP
14
gahhh i was writing (1-i)(1+i) as root 2
Reply 9
Avatar for kikzen
kikzen
OP
14
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)
untitled.JPG10.0KB
Reply 10
Avatar for kikzen
kikzen
OP
14
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!
Reply 11
Avatar for Gauss
Gauss
11
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.
untitled.JPG13.3KB
Reply 12
Avatar for kikzen
kikzen
OP
14
ooh that looks a lot nicer than mine :smile: at least i was ALONG the right lines, eh!
Reply 13
Avatar for Gauss
Gauss
11
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.

Related discussions

Latest

Trending

Trending

Articles for you