Question

Question: Use the Cauchy-Riemann conditions to find out whether the functions in Problems 1.1 to 1.21 are analytic.

$\frac{2z+3}{z+2}$

Short answer

The function $\frac{2z+3}{z+2}$is analytic everywhere except at$z=-2$.

Step-by-step solution

Given information

The given function is$\frac{2z+3}{z+2}$.

Concept of Cauchy-Riemann conditions

For the complex function $f\left(z\right)=f(x+iy)=u(x,y)+iv(x,y)$, where $u(x,y)$is the real part and $v(x,y)$is the imaginary part, to be analytic the conditions are$\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}$and$\frac{\partial v}{\partial x}=-\frac{\partial u}{\partial y}$.

Substitute the value

Substitute $z=x+iy$in $\frac{2z+3}{z+2}$and simplify.

$$\begin{gathered} \frac{2z+2}{z+2}=\frac{2(x+iy)+3}{(x+iy)+2} \\ =\frac{(2x+3)2iy}{(x+2+iy} \end{gathered}$$

Multiply numerator and denominator by $(x+2)-iy$:

role="math" localid="1653041513978" $$\begin{gathered} \frac{2z+3}{z+2}=\frac{\left[(2x+3)+2iy\right]\left[(x+2)-iy\right]}{\left[(x+2)+iy\right]\left[(x+2)-iy\right]} \\ =\frac{\left[(2x+3)(x+2)-iy(2x+3)+2iy(x+2)-2i^{2}y2\right]}{{(x+2)}^2-i^2{y}^2} \\ =\frac{(2x^2+4x+3x+6)+i(-2xy-3y+2xy+4y)+2y^2}{{(x+2)}^2+y^2} \\ =\frac{(2x^2+2y^2+7x+6)+iy}{{(x+2)}^2+y^2} \end{gathered}$$

Simplify the equation further gives:

$\frac{(2x^{2+}2y^2+7x+6)+iy}{{(x+2)}^2+y^2}=\frac{(2x^2+2y^2+7x+6)}{{(x+2)}^2+y^2}+i\frac{y}{{(x+2)}^2+y^2}$

Hence, the real part of the given function is$u(x,v)=\frac{(2x^2+2y^2+7x+6)}{{(x+2)}^2+y^2}$and the imaginary part is$v(y,x)=\frac{y}{{(x+2)}^2+y^2}$.

Apply Cauchy-Riemann conditions

Substitute the values of $u$and $y$in localid="1653042186012" $\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}$and $\frac{\partial v}{\partial x}=-\frac{\partial u}{\partial y}$and simplify.

localid="1653042612737" $$\begin{gathered} \frac{\partial u}{\partial x}=\frac{\left[{(x+2)}^2+y^2\right](4x+7)-2(x+2)(2x^2+2y^2+7x+6)}{{\left[{(x+2)}^2+y^2\right]}^2} \\ =\frac{x^2-y^2+4x+4}{{\left[{(x+2)}^2+y^2\right]}^2} \\ \frac{\partial u}{\partial y}=\frac{\left[{(x+2)}^2+y^2\right]\left(4y\right)-(2x^2+2y^2+7x+6)}{{\left[{(x+2)}^2+y^2\right]}^2} \\ =\frac{2xy+4y}{{\left[{(x+2)}^2+y^2\right]}^2} \\ \frac{\partial v}{\partial x}=\frac{\left[{(x+2)}^2+y^2\right].0-2(x+2)y}{{\left[{(x+2)}^2+y^2\right]}^2} \\ =-\frac{(2xy+4y)}{{\left[{(x+2)}^2+y^2\right]}^2} \\ \frac{\partial v}{\partial y}=\frac{\left[{(x+2)}^2+y^2\right].1-2y.y}{{\left[{(x+2)}^2+y^2\right]}^2} \\ =\frac{x^2-y^2+4x+4}{{\left[{(x+2)}^2+y^2\right]}^2} \end{gathered}$$

Here,$\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}$and$\frac{\partial v}{\partial x}=-\frac{\partial u}{\partial y}$, that is, this function satisfy the Cauchy-Riemann condition except$z=-2$.