Question

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

role="math" localid="1658486330935" $\frac{\mathbf{1}}{\mathbf{z}\mathbf{+}\mathbf{1}}$.

Short answer

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

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

Step-by-step solution

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

Find the value 

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

Find the values for 2-3i .

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

Find the values for $x+iy$.

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

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

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