Question

Find the real and imaginary parts $\mathit{u}\left(x,y\right)$and $\mathbf{v}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{)}$of the following functions.

$\frac{\mathbf{1}}{\mathbf{z}}$

Short answer

The real part$u\left(x,y\right)$of the function is $\frac{x}{x^2+y^2}$and the imaginary part$v\left(x,y\right)$of the function is$\frac{y}{x^2+y^2}$ .

Step-by-step solution

Definition of complex number

Complex number is represented by, where$\mathit{a}\mathbf{+}\mathit{i}\mathit{b}$is the a real number and bis the imaginary number.

Solve complex number

Given the function is $\frac{1}{z}$.

The complex number z can be written as:$\frac{1}{x+iy}$

$z=x+iy$, where x is a real part and y is an imaginary part.

The function $\frac{1}{z}$can be written as

Multiply with complex conjugate of $z=x+iy$into denominator and numerator to separate real and imaginary parts.

$$\begin{gathered} \frac{1}{z}=\frac{\left(x-iy\right)}{\left(x+iy\right)\left(x-iy\right)} \\ =\frac{\left(x-iy\right)}{x^2-i^2{y}^2} \\ =\frac{\left(x+iy\right)}{x^2+y^2} \end{gathered}$$

Find real and imaginary parts

Simplify the expression future

$\frac{\left(x+iy\right)}{x^2+y^2}=\frac{x}{x^2+y^2}+i\frac{y}{x^2+y^2}$

Hence the real part is $\frac{x}{x^2+y^2}$and imaginary part is $\frac{y}{x^2+y^2}$.