Q23P - Mathematical Methods in Physical Sciences

Question

Find each of the following in rectangular $\mathit{a}\mathbf{+}\mathit{z}\mathit{b}$ form if  $\mathit{z}\mathbf{=}\mathbf{2}\mathbf{-}{}_{\mathbf{t}}\mathbf{3}$ ; if $\mathit{z}\mathbf{=}\mathit{x}\mathbf{+}{}_{\mathbf{z}}\mathit{y}$ .

$\frac{\mathbf{1}\mathbf{+}\mathbf{z}}{\mathbf{1}\mathbf{-}\mathbf{z}}$.

Step-by-Step Solution

Answer

The value for $\frac{1+z}{1-z}$  is given below:

$$\begin{gathered} \frac{1+z}{1-z}=\frac{-6-3i}{5} \\ \frac{1+z}{1-z}=\frac{\left(1-x^2\right)+2yi-y^2}{\left(1-x^2\right)+y^2} \end{gathered}$$

 

Step 1: Given Information

The complex number is $\frac{1+z}{1-z}$.

Step 2: Definition of the Complex number

Complex number z can be written in the form of the real and imaginary numbers as:

$\mathit{z}\mathbf{=}\mathit{a}\mathbf{+}{}_{\mathbf{z}}\mathit{b}$
 

Here z is the complex number, and a and b are real numbers.

Step 3: Find the value.

$\frac{1+z}{1=z}$The complex number is $\frac{1+z}{1-z}$ .

 

Find the values for 2-3i .

$$\begin{gathered} \frac{1+z}{1-z}=\frac{3-3i}{-1+3i} \\ =\frac{3-3i}{-1+3i}\times \frac{-1-3i}{-1-3i} \\ =\frac{-3-9i+3i\left(3i^2\right)}{1-{\left(3i\right)}^2} \\ =\frac{-6-3i}{5} \end{gathered}$$

  

Find the values for $x+iy$ .

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

 

 

The value for $\frac{1+z}{1-z}$  is given below:

$$\begin{gathered} \frac{1+z}{1-z}=\frac{-6-3i}{5} \\ \frac{1+z}{1-z}=\frac{\left(1-x^2\right)+2yi-y^2}{{\left(1-x\right)}^2+y^2} \end{gathered}$$