QUESTION 6 Let f: ℂ→ℂ be defined by f(z) = sin z. (6.1)...

QUESTION 6 Let $f: \mathbb{C} \to \mathbb{C}$ be defined by $f(z) = \sin z$. (6.1) Write down an argument which demonstrates that $f(z) = 0$, if $z = n\pi$ ($n \in \mathbb{Z}$). (2) (6.2) Let $z = x + iy \in \mathbb{C}$. Use the fact that $\sin z = \sin x \cosh y + i \cos x \sinh y$ to show that (5) $|\sin z|^2 = \sin^2 x + \sinh^2 y$. (Hint: $\cosh^2 y - \sinh^2 y = 1$, for every $y \in \mathbb{R}$) (6.3) Use (6.2) to show that $z = n\pi$ ($n \in \mathbb{Z}$), if $f(z) = 0$. (4) [11]

Transcript

00:01 In this question, we have to prove that first of all, the function, which is a f -of -z is equals to e -raised -i -z, is an entire function.

00:13 Okay, so let us see how we are going to do this.

00:16 So, we know that e -r -r -z can be written in the form of series as i -t times z -rase to n divided by n factorial.

00:28 So here a and i'm considering as i z raise to n which can be written as i raise to n z so i will try to find out this limit l which is limit and tending to infinity under root of a n so which will give you i raise to n divided by n factorial correct? so what is this value? this is nothing but limit and tending to infinity under root of 1 over n factorial and this obviously is tending towards zero.

01:02 So if i write down the radius of convergence, capital r, which is 1 over l, so it will be 1 over 0, so it comes out to be infinity.

01:10 Okay, so hence for every epsilon positive, what do we get? we get 1 over under root of n factorial is obviously less than under root of n square, which is nothing but 1 over n, which is less than epsilon.

01:29 Okay, and here one thing to note that n is always greater than n -not, which we are considering as one over epsilon, correct? because why i'm writing this, so this part is because n factorial is greater than n squared and it is true for n greater or equals to 8, it holds.

01:51 Okay, so this implies that f of z is an entire, function now let's move to the next part in the next part what we have been given the function has no so we have this function erase to z has no zero in the field of complex number so how we are going to prove this so we will take if z is equals to z not let's say is a zero of this function f of z then what do we get then e -raise to z -0 times he -raise to minus z -0 which is nothing but z -0 minus z -0 minus of z -0 which is e -raised to 0 will be equals to 1 but but it is given that e -raise to z -not times he -rays to minus z -0 and e -raise to z -0 is a 0 so it will be 0 times he -rays to minus z -not is equals to 0 so this will imply that 0 is equal to 1 but this is is a contradiction okay why this is equals to zero since z not is a zero okay so we are getting this to be zero but then what we are getting zero is equal to one which is a contradiction so this implies that f of z is equals to erase to z as no zero in the field of complex number correct now we move to the next part in the next part what we have been given is the function erase to i theta which is equals to we know cosine of theta plus i times sine of theta with theta belongings to set of real numbers and the technometric function on the real line okay so how to do this question so we know first of all cosine theta has a series and going from zero to infinity minus 1 raised 2 n in terms of theta if i write it will be theta raised to 2n divided by 2 times of n factorial okay and sign of theta is what it is summation minus 1 raise to n theta raised to 2n plus 1 divided by 2n plus 1 factorial, correct? so if i write down cosine theta plus i time terms i times theta, so it will be the addition of these two.

04:36 So adding what we will get summation and going from 0 to infinity, i will get i theta raised to n divided by n factorial, which is same as e -raged to i theta.

04:49 So basically we needed to show that this two are equivalent.

04:53 So this is how we can show this, correct? now we move to the fourth part...

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