# How to show that e.g. $\cos(z)$ is analytic using Cauchy- Riemann differential equations?

How to show that e.g. $\cos(z)$ is analytic using Cauchy-Riemann differential equations [$u_x(x,y)=v_y(x,y)$ and $u_y(x,y)=-v_x(x,y)$]? Do all analytic functions satisfy Cauchy-Riemann differential equations (CRDE)? What is the relationship between analyticity of complex functions and Cauchy-Riemann differential equations? I know that holomorphic (analytic?) functions satisfy CRDE, but are functions that satisfy CRDE always analytic (holomorphic)?

## 5 Answers

Start by rewriting: if $z$ is complex, then let $z=x+iy$. Then we have the function $\cos(x+iy)$. Now you can expand that with the rule $\cos(A+B)=\cos(A)\cos(B)-\sin(A)\sin(B)$. (Because you'll be left with bits like e.g. $\cos(iy)$, you'll may want to replace these with hyperbolic functions with e.g. $\cos(ip)=\cosh(p)$ and a similar relationship for $\sin$.) You'll be left with some complex function which we'll call $u+vi$ - i.e. let $u$ be the real part and $v$ the imaginary part. It is these $u$ and $v$ you are differentiating in the Cauchy-Riemann equations.

• So if function is analytic( ~ holomorphic) in $\Omega \subset C$, then it satisfies C-R equations. And if f satisfies C-R equations and the functions $u(x,y)$ and $v(x,y)$ have first partial derivatives which are continuous, then $f$ is analytic(~holomorphic). Am I right? Then what about relationship between real differentiability and complex differentiability? Does they need C-R equations to distinguish them from each other? I mean that the relationship between those properties in case of complex function are defined by C-R equations . – user2723 Aug 23 '12 at 20:48

Writing $$z=x+iy\,\,,\,\,x,y\in\Bbb R\Longrightarrow \cos z=\frac{e^{iz}+e^{-iz}}{2}=\frac{\cos x(e^y+e^{-y})-i\sin x(e^y-e^{-y})}{2}=$$ $$\cos x\cosh y-i\sin x\sinh y=u+iv$$

And now you can check the Cauchy-Riemann equations directly, for example: $$u_x=-\sin x\cosh y=v_y\,\,,\,etc.$$

I will address your more general questions first. I'm restricting discussion to complex functions of one complex variable, of course.

Holomorphic and analytic functions are the same thing. A full proof is given in any complex analysis book, but I will give the outline. Assume all functions are defined in an open connected domain. Call functions with power series expansions at every point in their domain analytic, and call functions that complex-diferentiable holomorphic. If a function is analytic, we can expand it as a power series at every point, and elementary theory of power series shows they are complex-differentiable, so all analytic functions are holomorphic. To show all holomorphic functions are analytic, one uses a result called Cauchy's Integral Theorem to explicitly produce the required power series expansions at every point.

It is easily proved (see any book) that all holomorphic (complex-differentiable) functions satisfy the C-R equations, even without showing that holomorphic and analytic functions are the same. However, not all functions satisfying the Cauchy-Riemann equations are analytic (holomorphic). The standard additional condition is to require is that the function have continuous first partial derivatives (when considered as a function $\mathbb{R}^2 \rightarrow \mathbb{R}^2$ ) in addition to satisfying the C-R equations.

Now, showing $\cos(z)$ is analytic requires knowing how you defined it. I like to define it in terms of exponentials or power series, and in either case, analyticity is trivial. However, if you are restricted to the fact that $\cos(z)$ is just some function that satisfies the standard trigonometric properties, then I would go with the approach of Erik Pan, also posted in the answers section. Briefly: rewrite $\cos(x+iy)$ using the cosine addition formula, rewrite the result as $u+iv$ using hyperbolic functions, where $u$ and $v$ are real-valued, and then verify the C-R equations by differentiating directly. Remember to check that the first partial derivatives are continuous.

It sounds like you would benefit greatly from a good complex analysis book. A quick treatment (that covers your questions in greater detail than this answer) is available for free here.

If $f(z)$ is a complex-valued function, we can write it as $f(x+iy) = u(x,y) + iv(x,y)$, where $u$ and $v$ are real-valued functions.

Like you said, if $f$ is holomorphic, then the Cauchy-Riemann equations ($u_x = v_y$ and $u_y = -v_x$) are satisfied. However, the converse is not true. For example, if we let $f(x+iy) = \sqrt{|xy|}$, then $f$ satisfies the Cauchy-Riemann equations at $x=y=0$ but is not holomorphic there.

Thus, having $u_x = v_y$ and $u_y = -v_x$ at some point $z$ is not enough to conclude that $f$ is holomorphic there. However, if we add that $u$ and $v$ have continuous partial derivatives at $z$, then we can conclude that $f$ is holomorphic at $z$. (Alternatively, we could add that the mapping $(x,y) \mapsto (u(x,y), v(x,y))$ is differentiable as a $\mathbb{R}^2 \to \mathbb{R}^2$ function to conclude that $f$ is holomorphic at $x+iy$. This is a stronger result.)

In the case of $f(z) = \cos z$, we have $u(x,y) = \cos x\cosh y$ and $v(x,y) = -\sin x\sinh y$. We can check that the Cauchy Riemann equations hold everywhere, and furthermore, that all four partial derivatives are continuous everywhere. This is enough to conclude that $f$ is holomorphic.

$$f(z) = \cos z \implies u + iv = \cos(x + iy) = \cos x \cos iy - \sin x \sin iy = \cos x \cosh y - i \sin x \sinh y.$$ Here, $$u = \cos x \cosh y$$ ; $$v = - \sin x \sinh y$$ now, $$\frac{du}{dx} = - \sin x \cosh y$$ ; $$\frac{dv}{dx} = - \cos x \sinh y$$ ; $$\frac{du}{dy} = \cos x \sinh y$$ ; $$\frac{dv}{dy} = - \sin x \cosh y$$ We can see that, $$\frac{du}{dx} = \frac{dv}{dy}$$ and $$\frac{du}{dy} = - \frac{dv}{dx}$$ so, Cauchy Riemann equations are satisfied. we clearly see that $$\frac{du}{dx}, \frac{dv}{dx}, \frac{du}{dy}, \frac{dv}{dy}$$ are continuous. so, the function f(z) is analytic everywhere.

• What's new in your answer compared with the other answers posted seven years ago? – José Carlos Santos Nov 20 '19 at 18:54