F(z) is a function of the complex variable z = x +iy given by

𝐹(𝑧) = 𝑖 𝑧 + 𝑘 𝑅𝑒(𝑧) + 𝑖 𝐼𝑚(𝑧).

For what value of k will F z( ) satisfy the Cauchy-Riemann equations?

The correct answer is 1.

Step-by-Step Explanation:

We are given a function of the complex variable $z=x+iy$ defined as:

$F\left(z\right)=iz+k\text{Re}\left(z\right)+i\text{Im}\left(z\right)$

First, we express $F\left(z\right)$ in terms of its real and imaginary parts. We know that:
$\text{Re}\left(z\right)=x$
$\text{Im}\left(z\right)=y$

Substituting these along with $z=x+iy$ into the expression for $F\left(z\right)$:
$F\left(z\right)=i(x+iy)+kx+iy$
$F\left(z\right)=ix+i^{2}y+kx+iy$

Since $i^2=-1$, we have:
$F\left(z\right)=ix-y+kx+iy$

Grouping the real and imaginary parts together:
$F\left(z\right)=(kx-y)+i(x+y)$

Thus, we can write $F\left(z\right)=u(x,y)+iv(x,y)$, where:
$u(x,y)=kx-y$
$v(x,y)=x+y$

As shown in the first attached image (image_0.png), a complex function satisfies the Cauchy-Riemann equations if and only if:
$\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}$ and $\frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}$

Let us compute the required partial derivatives:
$\frac{\partial u}{\partial x}=\frac{\partial }{\partial x}(kx-y)=k$
$\frac{\partial u}{\partial y}=\frac{\partial }{\partial y}(kx-y)=-1$
$\frac{\partial v}{\partial x}=\frac{\partial }{\partial x}(x+y)=1$
$\frac{\partial v}{\partial y}=\frac{\partial }{\partial y}(x+y)=1$

Now, we substitute these derivatives into the Cauchy-Riemann equations:

1. The second equation, $\frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}$, becomes:
$-1=-\left(1\right)$
This is always true and does not impose any condition on $k$.

2. The first equation, $\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}$, as highlighted in the second attached image (image_1.png), yields:
$k=1$

Thus, the function $F\left(z\right)$ satisfies the Cauchy-Riemann equations if and only if $k=1$.