 Hello, everyone. Myself, AS Falmari, Assistant Professor, Department of Humanities and Sciences, Balchand Institute of Technology, Sholapur. In the last video, we have defined analytic functions, necessary and sufficient conditions for the analytic functions, and we have also studied some important notes based on the Cauchy-Riemann equations and the analyticity. In this video, we will discuss analyticity and the derivative of some elementary trigonometric functions. The learning outcome of this lesson is, at the end of this lesson, students will be able to verify Cauchy-Riemann equations and analyticity of a given complex valued function. So, before going to start this lesson, pause this video and check whether the functions e raised to x sin x cos x hyperbolic sin x hyperbolic cos x are continuous. I hope all of you have returned an answer. All these functions are real valued differentiable functions and we are knowing the property that differentiability implies continuity. Therefore, as these functions are a differentiable one, that are also continuous. Therefore, the answer for the question is, all the above functions are continuous functions. Let us consider some examples. Example number one, show that the function e raised to x into cos y plus i sin y is analytic. Solution. Let us denote the given function by f of z and f of z has always in general form u plus iv and it is here provided as e raised to x into cos y plus i sin y. Let us multiply by the exponential to this bracket we get the function as e raised to x into cos y plus i e raised to x into sin y. Both the sides are of the complex expressions. Therefore, equating their real and imaginary parts, we get u equal to e raised to x into cos y and v equal to e raised to x into sin y. Now, in order to check whether the functions are analytic or not, we will use the sufficient condition for the analyticity. The first condition is to check whether the given function satisfies the Cauchy-Riemann equations or not. Let us consider the first one CR equations. Here, we have provided the functions u and v. Let us differentiate this function u partially with respect to x treating y constant we get dou u by dou x equal to as y is constant cos y is constant write it as it is and the derivative of exponential as it is e raised to x. Similarly, differentiate u partially with respect to y treating x constant as x is constant e raised to x as it is and we know that the derivative of cos y is minus sin y. Therefore, the derivative becomes minus e raised to x into sin y. Similarly, differentiating v partially with respect to x we get the derivative of exponential e raised to x as it is and sin y is constant we have to write it as it is and differentiating v partially with respect to y treating x constant we get as x is constant write e raised to x as it is and the derivative of sin y is cos y. Now, from these four derivatives we can see that dou u by dou x and dou v by dou y has the same expression it means dou u by dou x equal to dou v by dou y and dou u by dou y is equal to minus e raised to x into sin y and dou v by dou x is equal to e raised to x sin y. Now, there is a difference of the sin only therefore, we can say that dou u by dou y is equal to minus of dou v by dou x that is from these four equations we can see that the Cauchy-Riemann conditions are satisfied. Let us consider the second condition continuity of the all these four derivatives. Now, these are the corresponding four derivatives just we have discussed that exponential sin are the continuous real valued functions again we are knowing the theorem that the product of two continuous functions is again a continuous function. Here all the four partial derivatives we have obtained as the product of continuous real valued functions therefore, they are also continuous therefore, by the sufficient condition the given function f of z is analytic everywhere here we have not obtained any condition for the x and y therefore, we can say that f of z is analytic everywhere in the complex plane f of z is an entire function. Now, let us consider the second example test the analyticity of the function w equal to sin z and hence derive that the derivative of sin z as cos z solution. Let us denote the given function by f of z and we know that f of z has a in general form u plus i v and here it is provided as sin z. We know that z is a complex number which is given by z equal to x plus i y. So, we can write this z as x plus i y we get sin z as sin of x plus i y. Again we know that the trigonometric identity sin of a plus b equal to sin a into cos b plus cos a into sin b by this identity the sin of x plus i y reduces to sin x into cos of i y plus cos x into sin of i y. Again we are knowing the relation between the trigonometric and hyperbolic functions cos of i y can be replaced by hyperbolic cos y and sin of i y can be replaced by i hyperbolic sin y. After using these two identities in this equation we get u plus i v equal to sin x into hyperbolic cos y plus i into cos x into hyperbolic sin y. Now, both the sides are the complex functions. So, equating the real and imaginary parts we get u equal to sin x into hyperbolic cos y and we equal to cos x into hyperbolic sin y. Now, in order to check whether the given function is analytic or not we will use the sufficient condition for the analyticity. The first condition is to check the Cr equations. Differentiating u partially with respect to x treating y constant we get. Now, as y is constant we can write hyperbolic cos y as it is and the derivative of sin x is cos x. Again differentiating u partially with respect to y treating x constant. Now, as x is constant we can write sin x as it is and the derivative of hyperbolic cos y is hyperbolic sin y. Similarly differentiating we partially with respect to x treating y constant we get. As y is constant we can write hyperbolic sin y as it is and the derivative of cos x is minus sin x. And finally differentiating we partially with respect to y treating x constant. As x is constant we can write cos x as it is and the derivative of hyperbolic sin y is hyperbolic cos y. From these four equations now here we can see that dou u by dou x is exactly equal to dou v by dou y and this dou u by dou y is nothing but minus of dou v by dou x that is what here we have written. From these two equations we can say that the given functions u and v satisfies these CR equations. Now let us consider the second condition continuity of all these four partial derivatives. Now here we can see that these partial derivative contains the product of sin cos and the hyperbolic sin and the hyperbolic cos. And we are knowing that the trigonometric sin cos and hyperbolic sin cos are the continuous functions. Again we are knowing that the product of two continuous functions is continuous so that we can say that all the four partial derivatives are continuous. Therefore the two two sufficient conditions are satisfied hence we can say that the given function f of z equal to sin z is analytic everywhere that is the f of z is an entire function. Now to obtain its derivative in the last video we have studied the theorem that the derivative of an analytic function is given as f dash of z equal to ux that is ux here represents the partial derivative of u with respect to x plus i into partial derivative of v with respect to x is equal to. Here we can see that ux is nothing but cos x into hyperbolic cos y and this vx is nothing but vx is nothing but minus sin x into hyperbolic sin y. Now in order to convert it to the function of z we will transform these hyperbolic functions into trigonometric with the help of the relation between trigonometric and hyperbolic functions. Now we are knowing that hyperbolic cos y equal to cos of iy let us replace this hyperbolic cos y by cos of iy and minus as it is and replace i hyperbolic sin y by sin iy. Now we can see that this is the expansion of cos of x plus iy which is nothing but cos a cos b minus sin a sin b and we are knowing that x plus iy is nothing but what z therefore cos of x plus iy can be written as cos z. Now this is the derivative of sin z. One important note similarly we can show that the derivative of the function f of z equal to cos z is minus sin z.