The value of k that makes the complex-valued function

𝑓(𝑧) = 𝑒 ^−𝑘𝑥 (cos 2𝑦 − 𝑖 sin 2𝑦) analytic,

where 𝑧 = 𝑥 + 𝑖𝑦, is _________.

(Answer in integer)

The correct answer is 2.

Step-by-step Explanation:

A complex-valued function $f\left(z\right)=u(x,y)+iv(x,y)$ is analytic if it satisfies the Cauchy-Riemann equations:

1) $\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}$

2) $\frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}$

From the given function in the problem and the details shown in the image:
$f\left(z\right)=e^{-kx}\left(\cos \right(2y)-i\sin (2y\left)\right)=e^{-kx}\cos \left(2y\right)-i{e}^{-kx}\sin \left(2y\right)$

We identify the real part $u(x,y)$ and the imaginary part $v(x,y)$ as follows:
$u(x,y)=e^{-kx}\cos \left(2y\right)$
$v(x,y)=-e^{-kx}\sin \left(2y\right)$

Next, we calculate the partial derivatives with respect to $x$ and $y$:

• Differentiating $u$ with respect to $x$:
$\frac{\partial u}{\partial x}=-k{e}^{-kx}\cos \left(2y\right)$

• Differentiating $u$ with respect to $y$:
$\frac{\partial u}{\partial y}=-2e^{-kx}\sin \left(2y\right)$

• Differentiating $v$ with respect to $x$:
$\frac{\partial v}{\partial x}=k{e}^{-kx}\sin \left(2y\right)$

• Differentiating $v$ with respect to $y$:
$\frac{\partial v}{\partial y}=-2e^{-kx}\cos \left(2y\right)$

Now we substitute these derivatives into the first Cauchy-Riemann equation:
$\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}$

$-k{e}^{-kx}\cos \left(2y\right)=-2e^{-kx}\cos \left(2y\right)$

Equating the coefficients on both sides, we get:
$k=2$

Let us verify if this satisfies the second Cauchy-Riemann equation:
$\frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}$

$-2e^{-kx}\sin \left(2y\right)=-\left(k{e}^{-kx}\sin \right(2y\left)\right)$

Substituting $k=2$ yields:
$-2e^{-2x}\sin \left(2y\right)=-2e^{-2x}\sin \left(2y\right)$

Both equations are satisfied when $k=2$. Therefore, the value of $k$ that makes the function analytic is 2.