Question

Find real $\mathit{x}$and $\mathit{y}$for which$\left|z+3\right|\mathbf{=}\mathbf{1}\mathbf{-}\mathit{i}\mathit{z}$, where $\mathit{z}\mathbf{=}\mathit{x}\mathbf{+}\mathit{i}\mathit{y}$.

Short answer

The real values of $\left(x,y\right)=\left(0,4\right)$.

Step-by-step solution

Given Information

The given equation is $\left|z+3\right|=1-iz$.

Definition of Complex Numbers

The numbers that are presented in the form of $\mathit{x}\mathbf{+}\mathit{i}\mathit{y}$where, $\mathbf{\text{'}}\mathit{x}\mathbf{\text{'}}$is real numbers and $\mathbf{\text{'}}\mathit{i}\mathit{y}\mathbf{\text{'}}$ is an imaginary number, those numbers are referred to as called Complex numbers.

 Find the ratio of the series.

Consider, $\left|z+3\right|=1-iz$. …(1)

Let $z=x+iy$and put in equation (1).

$$\begin{gathered} \left|x+yi+3\right|=1-\left(x-yi\right) \\ \left|x+yi+3\right|=\left(1+y\right)-xi \end{gathered}$$ ..(2)

Compare both the real and imaginary parts of the equation (2).

The real part is $\left|x+yi+3\right|=\left(1+y\right)$. ....(3)

The imaginary part is $-xi=0$. …(4)

Use equation (4) to get the value of $x=0$.

Put the value $x=0$in equation (3).

$$\begin{gathered} \left|yi+3\right|=\left(1+y\right) \\ \left|yi+3\right|=\sqrt{y^2+3^2} \end{gathered}$$

Compare both the above equation.

$$\begin{gathered} \sqrt{y^2+3^2}=1+y \\ {y}^2+3^2={\left(1+y\right)}^2 \\ {y}^2+3^2=1+2y+y^2 \\ y=4 \end{gathered}$$

Hence the real values of $\left(x,y\right)=\left(0,2\right)$.