Question

Solve the equation:
$\left| z \right|+z=2+i$

Answer 1

Hint: Let us assume that $z=x+iy$ then $ | z |=\sqrt{{{x}^{2}}+{{y}^{2}}}$. Substitute the values of z and $\left| z \right|$ in the given equation and then equate left hand side and right hand side of the given equation. After equating L.H.S and R.H.S of the given equation, you will get the value of complex number z in terms of x and y.

Complete step-by-step solution -
The equation that we are going to solve is:
 $\left| z \right|+z=2+i$
In the above equation, let us assume that $z=x+iy$.
Then from the complex number, we know that:
$\left| z \right|=\sqrt{{{x}^{2}}+{{y}^{2}}}$
Now, substituting the values of z and $\left| z \right|$ in the equation given in the question we get,
$\begin{align}
  & \left| z \right|+z=2+i \\
 & \Rightarrow \sqrt{{{x}^{2}}+{{y}^{2}}}+x+iy=2+i \\
\end{align}$
Equating the real part of the left hand side with the real part of the right hand side of the equation we get,
$\begin{align}
  & \sqrt{{{x}^{2}}+{{y}^{2}}}+x=2 \\
 & \Rightarrow \sqrt{{{x}^{2}}+{{y}^{2}}}=2-x \\
\end{align}$
Squaring on both the sides of the above equation will give:
$\begin{align}
  & {{x}^{2}}+{{y}^{2}}={{\left( 2-x \right)}^{2}} \\
 & \Rightarrow {{x}^{2}}+{{y}^{2}}=4+{{x}^{2}}-4x \\
\end{align}$
From the above equation, we can see that ${{x}^{2}}$ will be cancelled out from both the sides.
${{y}^{2}}=4-4x$……… Eq. (1)
In this equation $\sqrt{{{x}^{2}}+{{y}^{2}}}+x+iy=2+i$, we are going to equate the imaginary part of both left and right hand side.
$y=1$
Substituting this value of y in equation (1) we get,
$\begin{align}
  & {{y}^{2}}=4-4x \\
 & \Rightarrow 1=4-4x \\
 & \Rightarrow 4x=3 \\
 & \Rightarrow x=\dfrac{3}{4} \\
\end{align}$
Substituting the values of x and y in $z=x+iy$ that we have obtained from above we get,
$z=\dfrac{3}{4}+i$

Note: Some properties of modulus of complex number z:
$\left| z \right|$ is the magnitude of the vector z in the argand plane.
$\left| z \right|=\sqrt{{{x}^{2}}+{{y}^{2}}}$
As $\left| z \right|$ is the modulus of z so it cannot be negative.
You might think of squaring both the sides in the given equation $\left| z \right|+z=2+i$ and then solve.
After squaring on both the sides of this equation, you will find z and $\left| z \right|$ terms on the left hand side of the equation then it is very difficult to equate real and imaginary parts on both sides of the equation because the right hand side contains the term “i” but the left hand side contains no “i” term.