Question

If $|Z_1+Z_2|$ = $\left|Z_1\right|+\left|Z_2\right|$, then find the value of arg $\left(\frac{Z_1}{Z_2}\right)$

• A.
• B.
• C.
• D. 0

Answer 1

The correct option is D

0

arg $\left(\frac{Z_1}{Z_2}\right)$ is the angle between $\left|Z_1\right|$ and $\left|Z_2\right|$

In the figure, we see that modulus of $Z_1$+ $Z_2$ is equal to the sum of modulus of $Z_1$ and $Z_2$.

The angle between $Z_1$ and $Z_2$ is Zero.
Another way of solving is by algebraically in the following way :
${(\left|Z_1\right|+\left|Z_2\right|)}^2={|Z_1+Z_2|}^2=(Z_1+Z_2)(\overline{Z_1}+\overline{Z_2})$
${\left|Z_1\right|}^2+{\left|Z_2\right|}^2+2\left|Z_1\right|\left|Z_2\right|=Z_1\overline{Z_1}+Z_2\overline{Z_2}+Z_1\overline{Z_2}+Z_2\overline{Z_1}$
= ${\left|Z_1\right|}^2+{\left|Z_2\right|}^2+Z_1\overline{Z_2}+\overline{Z_1\overline{Z_2}}$
${\left|Z_1\right|}^2+{\left|Z_2\right|}^2+2\left|Z_1\right|\left|Z_2\right|={\left|Z_1\right|}^2+{\left|Z_2\right|}^2+2Re\left(Z_1\overline{Z_2}\right)$
$\Rightarrow$ |$Z_1$| |$Z_2$|=Re($Z_1$${\overline{Z}}_2$) ------ (1)
Let $Z_1$=r $e^{ia}$ and $Z_2$ = r $e^{iA}$
Substituting in (1), we get
R.r = Re(r $e^{ia}$, R $e^{iA}$)
Re(r.R$e^{i(a-A)}$)
Re(r.R$e^{i(a-A)}$) to be equal to R.r, we should have the real part of $e^{i(a-A)}$ to be equal to one.
This is possible when A=a.
So, the arguement of both $Z_1$ and $Z_2$ should be equal or the angle between them is zero.
$\Rightarrow$ $Z_1$ and $Z_2$ are parallel