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?

To find the value of $k$ for which the function $F\left(z\right)$ satisfies the Cauchy-Riemann equations, we first express $F\left(z\right)$ in terms of its real and imaginary parts.

We are given:
$z=x+iy$
where $x=\mathrm{Re}\left(z\right)$ and $y=\mathrm{Im}\left(z\right)$.

The function is given by:
$F\left(z\right)=iz+k\mathrm{Re}\left(z\right)+i\mathrm{Im}\left(z\right)$

Substitute $z=x+iy$, $\mathrm{Re}\left(z\right)=x$, and $\mathrm{Im}\left(z\right)=y$ into the expression:
$F\left(z\right)=i(x+iy)+kx+iy$
$F\left(z\right)=ix-y+kx+iy$

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

Thus, the real part $u(x,y)$ and the imaginary part $v(x,y)$ are:
$u(x,y)=kx-y$
$v(x,y)=x+y$

For $F\left(z\right)$ to satisfy the Cauchy-Riemann equations, the following partial derivatives must be equal:
1) $\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}$
2) $\frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}$

Let's calculate the partial derivatives:
$\frac{\partial u}{\partial x}=\frac{\partial }{\partial x}(kx-y)=k$
$\frac{\partial u}{\partial y}=\frac{\partial }{\partial y}(kx-y)=-1$
$\frac{\partial v}{\partial x}=\frac{\partial }{\partial x}(x+y)=1$
$\frac{\partial v}{\partial y}=\frac{\partial }{\partial y}(x+y)=1$

Checking the second equation:
$\frac{\partial u}{\partial y}=-1$ and $-\frac{\partial v}{\partial x}=-1$
This equation is satisfied for any value of $k$.

Checking the first equation:
$\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}\Rightarrow k=1$

Therefore, $F\left(z\right)$ satisfies the Cauchy-Riemann equations when $k=1$.