Question

If ${{z}_{1}}$ and ${{z}_{2}}$ are two complex numbers, then
A. $\left| {{z}_{1}}+{{z}_{2}} \right|\ge \left| {{z}_{1}} \right|+\left| {{z}_{2}} \right|$
B. $\left| {{z}_{1}}-{{z}_{2}} \right|\ge \left| {{z}_{1}} \right|+\left| {{z}_{2}} \right|$
C. $\left| {{z}_{1}}-{{z}_{2}} \right|\le \left| {{z}_{1}} \right|-\left| {{z}_{2}} \right|$
D. $\left| {{z}_{1}}+{{z}_{2}} \right|\ge \left| {{z}_{1}} \right|-\left| {{z}_{2}} \right|$

Answer 1

Hint: In this question, we are to prove the given property of the complex numbers. For this, the basic operations are applied.

Formula used: The complex number $(x,y)$ is represented by $x+iy$.
If $z=x+iy\in C$, then $x$ is called the real part and $y$ is called the imaginary part of $z$. These are represented by $\operatorname{Re}(z)$ and $\operatorname{Im}(z)$ respectively.
$z=x+iy$ be a complex number such that $\left| z \right|=r$ and $\theta $ be the amplitude of $z$. So, $\cos \theta =\dfrac{x}{r},\sin \theta =\dfrac{b}{r}$
And we can write the magnitude as
 $\begin{align}
  & \left| z \right|=\left| x+iy \right| \\
 & \Rightarrow r=\sqrt{{{x}^{2}}+{{y}^{2}}} \\
\end{align}$
Thus, we can write
$z=x+iy=r\cos \theta +ir\sin \theta =r(\cos \theta +i\sin \theta )$
This is said to be the mod amplitude form or the polar form of $z$.
Where $\cos \theta +i\sin \theta $ is denoted by $cis\theta $ and the Euler’s formula is $\cos \theta +i\sin \theta ={{e}^{i\theta }}$
Some of the basic properties of complex numbers are:
$\operatorname{Re}(z)\le \left| z \right|,\operatorname{Im}(z)\le \left| z \right|$
$\begin{align}
  & \overline{{{z}_{1}}+{{z}_{2}}}=\overline{{{z}_{1}}}+\overline{{{z}_{2}}} \\
 & \overline{{{z}_{1}}-{{z}_{2}}}=\overline{{{z}_{1}}}-\overline{{{z}_{2}}} \\
 & \overline{{{z}_{1}}{{z}_{2}}}=\overline{{{z}_{1}}}\overline{{{z}_{2}}} \\
 & \left( \overline{{}^{{{z}_{1}}}/{}_{{{z}_{2}}}} \right)={}^{\overline{{{z}_{1}}}}/{}_{\overline{{{z}_{2}}}};{{z}_{2}}\ne 0 \\
\end{align}$

Complete step by step solution: Given that, ${{z}_{1}}$ and ${{z}_{2}}$ are two complex numbers.
Consider $\left| {{z}_{1}} \right|>\left| {{z}_{2}} \right|$
Then, we can write
\[\left| {{z}_{1}} \right|=\left| ({{z}_{1}}+{{z}_{2}})-{{z}_{2}} \right|\text{ }...(1)\]
But we know that,
$\left| {{z}_{1}}-{{z}_{2}} \right|\le \left| {{z}_{1}} \right|+\left| {{z}_{2}} \right|$
Then, on applying the above property in (1), we get
\[\begin{align}
  & \left| {{z}_{1}} \right|=\left| ({{z}_{1}}+{{z}_{2}})-{{z}_{2}} \right|\le \left| {{z}_{1}}+{{z}_{2}} \right|+\left| -{{z}_{2}} \right| \\
 & \Rightarrow \left| {{z}_{1}} \right|\le \left| {{z}_{1}}+{{z}_{2}} \right|+\left| {{z}_{2}} \right| \\
 & \Rightarrow \left| {{z}_{1}} \right|-\left| {{z}_{2}} \right|\le \left| {{z}_{1}}+{{z}_{2}} \right| \\
 & \Rightarrow \left| {{z}_{1}}+{{z}_{2}} \right|\ge \left| {{z}_{1}} \right|-\left| {{z}_{2}} \right|\text{ }...(2) \\
\end{align}\]
If $\left| {{z}_{2}} \right|>\left| {{z}_{1}} \right|$
Then, we can write
\[\left| {{z}_{2}} \right|=\left| ({{z}_{1}}+{{z}_{2}})-{{z}_{1}} \right|\text{ }...(3)\]
But we know that,
$\left| {{z}_{1}}-{{z}_{2}} \right|\le \left| {{z}_{1}} \right|+\left| {{z}_{2}} \right|$
Then, on applying the above property in (1), we get
\[\begin{align}
  & \left| {{z}_{2}} \right|=\left| ({{z}_{1}}+{{z}_{2}})-{{z}_{1}} \right|\le \left| {{z}_{1}}+{{z}_{2}} \right|+\left| -{{z}_{1}} \right| \\
 & \Rightarrow \left| {{z}_{2}} \right|\le \left| {{z}_{1}}+{{z}_{2}} \right|+\left| {{z}_{1}} \right| \\
 & \Rightarrow \left| {{z}_{2}} \right|-\left| {{z}_{1}} \right|\le \left| {{z}_{1}}+{{z}_{2}} \right| \\
 & \Rightarrow \left| {{z}_{1}}+{{z}_{2}} \right|\ge \left| {{z}_{2}} \right|-\left| {{z}_{1}} \right|\text{ }...(4) \\
\end{align}\]
Thus, from (3) and (4), we get
$\left| {{z}_{1}}+{{z}_{2}} \right|\ge \left| \left| {{z}_{1}} \right|-\left| {{z}_{2}} \right| \right|$

Thus, Option (D) is correct.

Note: Here we need to apply the properties of complex numbers, to find the given expression. By applying appropriate formulae, the required value is obtained.