Question

Write and Explain The Cauchy Riemann Equation

Answer 1

Answer:ok va bro

Step-by-step explanation:

Formally we may write the derivative of {\displaystyle f(z)}{\displaystyle f(z)} with respect to {\displaystyle z}{\displaystyle z} using the classical definition of the derivative

{\displaystyle {\frac {df(z)}{dz}}=\lim _{\Delta z\rightarrow 0}{\frac {f(z+\Delta z)-f(z)}{\Delta z}}.}{\displaystyle {\frac {df(z)}{dz}}=\lim _{\Delta z\rightarrow 0}{\frac {f(z+\Delta z)-f(z)}{\Delta z}}.}

Here, of course, we recognize that the theory of limits of complex valued sequences and series can be established.

If we write {\displaystyle f(z)}{\displaystyle f(z)} in terms of its real and imaginary parts, the formal representation of the derivative becomes

{\displaystyle {\frac {df(z)}{dz}}=\lim _{\Delta x\rightarrow 0,\;\Delta y\rightarrow 0}{\frac {u(x+\Delta x,y+\Delta y)-u(x,y)+i(v(x+\Delta x,y+\Delta y)-v(x,y))}{\Delta x+i\Delta y}}}{\displaystyle {\frac {df(z)}{dz}}=\lim _{\Delta x\rightarrow 0,\;\Delta y\rightarrow 0}{\frac {u(x+\Delta x,y+\Delta y)-u(x,y)+i(v(x+\Delta x,y+\Delta y)-v(x,y))}{\Delta x+i\Delta y}}}

where {\displaystyle \Delta z=\Delta x+i\Delta y.}{\displaystyle \Delta z=\Delta x+i\Delta y.}

For the derivative to exist, the limit must exist and be unique. Therefore, it should not matter whether we take the {\displaystyle \lim _{\Delta x\rightarrow 0}}{\displaystyle \lim _{\Delta x\rightarrow 0}} first or then {\displaystyle \lim _{\Delta y\rightarrow 0}}{\displaystyle \lim _{\Delta y\rightarrow 0}} second, or vice versa. We must obtain an equivalent result either way.

If we take the {\displaystyle \lim _{\Delta y\rightarrow 0}}{\displaystyle \lim _{\Delta y\rightarrow 0}} first, then the resulting limiting process is

{\displaystyle {\frac {df(z)}{dz}}=\lim _{\Delta x\rightarrow 0}{\frac {u(x+\Delta x,y)-u(x,y)+i(v(x+\Delta x,y)-v(x,y))}{\Delta x}}.}{\displaystyle {\frac {df(z)}{dz}}=\lim _{\Delta x\rightarrow 0}{\frac {u(x+\Delta x,y)-u(x,y)+i(v(x+\Delta x,y)-v(x,y))}{\Delta x}}.}

Recognizing the partial derivative with respect to {\displaystyle x}{\displaystyle x} we have

{\displaystyle {\frac {df(z)}{dz}}={\frac {\partial u(x,y)}{\partial x}}+i{\frac {\partial v(x,y)}{\partial x}}.}{\displaystyle {\frac {df(z)}{dz}}={\frac {\partial u(x,y)}{\partial x}}+i{\frac {\partial v(x,y)}{\partial x}}.}

Alternatively, we may take the {\displaystyle \lim _{\Delta x\rightarrow 0}}{\displaystyle \lim _{\Delta x\rightarrow 0}} and the resulting limit becomes

{\displaystyle {\frac {df(z)}{dz}}=\lim _{\Delta y\rightarrow 0}{\frac {u(x,y+\Delta y)-u(x,y)+i(v(x,y+\Delta y)-v(x,y))}{i\Delta y}}={\frac {\partial v(x,y)}{\partial y}}-i{\frac {\partial u(x,y)}{\partial y}}.}{\displaystyle {\frac {df(z)}{dz}}=\lim _{\Delta y\rightarrow 0}{\frac {u(x,y+\Delta y)-u(x,y)+i(v(x,y+\Delta y)-v(x,y))}{i\Delta y}}={\frac {\partial v(x,y)}{\partial y}}-i{\frac {\partial u(x,y)}{\partial y}}.}

Thus, we have found two ways to represent {\displaystyle df(z)/dz}{\displaystyle df(z)/dz}.

Because both of these expressions for {\displaystyle df/dz}{\displaystyle df/dz} must be equivalent, we equate the real and imaginary parts of these expressions to obtain

{\displaystyle {\frac {\partial u(x,y)}{\partial x}}={\frac {\partial v(x,y)}{\partial y}}\quad \quad {\mbox{and}}\quad \quad {\frac {\partial v(x,y)}{\partial x}}=-{\frac {\partial u(x,y)}{\partial y}}.}{\displaystyle {\frac {\partial u(x,y)}{\partial x}}={\frac {\partial v(x,y)}{\partial y}}\quad \quad {\mbox{and}}\quad \quad {\frac {\partial v(x,y)}{\partial x}}=-{\frac {\partial u(x,y)}{\partial y}}.}