Question

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

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.
So, we have
$\operatorname{Re}(z)\le \left| z \right|,\operatorname{Im}(z)\le \left| z \right|$
If $z=x+iy\in C$ then $\overline{z}=x-iy$
On multiplying $z$ and $\overline{z}$, we get
$\begin{align}
  & z\overline{z}=(x+iy)(x-iy) \\
 & \text{ }={{x}^{2}}+{{y}^{2}} \\
 & \text{ }={{\left| z \right|}^{2}} \\
\end{align}$
Applying this for the given expression $\left| {{z}_{1}}+{{z}_{2}} \right|$
$\begin{align}
  & {{\left| {{z}_{1}}+{{z}_{2}} \right|}^{2}}=({{z}_{1}}+{{z}_{2}})(\overline{{{z}_{1}}+{{z}_{2}}}) \\
 & \text{ }=({{z}_{1}}+{{z}_{2}})(\overline{{{z}_{1}}}+\overline{{{z}_{2}}}) \\
 & \text{ }={{z}_{1}}\overline{{{z}_{1}}}+{{z}_{1}}\overline{{{z}_{2}}}+{{z}_{2}}\overline{{{z}_{1}}}+{{z}_{2}}\overline{{{z}_{2}}} \\
 & \text{ }={{\left| {{z}_{1}} \right|}^{2}}+{{\left| {{z}_{2}} \right|}^{2}}+({{z}_{1}}\overline{{{z}_{2}}}+{{z}_{2}}\overline{{{z}_{1}}})\text{ }...(1) \\
\end{align}$
For finding the term $({{z}_{1}}\overline{{{z}_{2}}}+{{z}_{2}}\overline{{{z}_{1}}})$, consider ${{z}_{1}}=a+ib;{{z}_{2}}=c+id$
$\begin{align}
  & ({{z}_{1}}\overline{{{z}_{2}}}+{{z}_{2}}\overline{{{z}_{1}}})=(a+ib)(c-id)+(a-ib)(c+id) \\
 & \text{ }=ac-iad+ibc+bd+ac+iad-ibc+bd \\
 & \text{ }=2(ac+bd) \\
 & \text{ }=2\operatorname{Re}({{z}_{1}}\overline{{{z}_{2}}}) \\
\end{align}$
Then, on substituting in (1), we get
${{\left| {{z}_{1}}+{{z}_{2}} \right|}^{2}}={{\left| {{z}_{1}} \right|}^{2}}+{{\left| {{z}_{2}} \right|}^{2}}+2\operatorname{Re}({{z}_{1}}\overline{{{z}_{2}}})\text{ }...(2)$
But we have $\operatorname{Re}(z)\le \left| z \right|$
So,
\[\begin{align}
  & \operatorname{Re}({{z}_{1}}\overline{{{z}_{2}}})\le \left| {{z}_{1}}\overline{{{z}_{2}}} \right|=\left| {{z}_{1}} \right|\left| \overline{{{z}_{2}}} \right|=\left| {{z}_{1}} \right|\left| {{z}_{2}} \right| \\
 & \Rightarrow \operatorname{Re}({{z}_{1}}\overline{{{z}_{2}}})\le \left| {{z}_{1}} \right|\left| {{z}_{2}} \right|\text{ }...(3) \\
\end{align}\]
Thus, from (2) and (3),
\[\begin{align}
  & {{\left| {{z}_{1}}+{{z}_{2}} \right|}^{2}}={{\left| {{z}_{1}} \right|}^{2}}+{{\left| {{z}_{2}} \right|}^{2}}+2\operatorname{Re}({{z}_{1}}\overline{{{z}_{2}}}) \\
 & \Rightarrow {{\left| {{z}_{1}}+{{z}_{2}} \right|}^{2}}\le {{\left| {{z}_{1}} \right|}^{2}}+{{\left| {{z}_{2}} \right|}^{2}}+2\left| {{z}_{1}} \right|\left| {{z}_{2}} \right| \\
 & \Rightarrow {{\left| {{z}_{1}}+{{z}_{2}} \right|}^{2}}\le {{\left( \left| {{z}_{1}} \right|+\left| {{z}_{2}} \right| \right)}^{2}} \\
 & \therefore {{\left| {{z}_{1}}+{{z}_{2}} \right|}^{2}}\le \left| {{z}_{1}} \right|+\left| {{z}_{2}} \right| \\
\end{align}\]

Thus, Option (A) 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.