Question

If \[{z_1}\] and \[{z_2}\] are two complex numbers that \[\left| {{z_1}} \right| = \left| {{z_2}} \right| + \left| {{z_1} - {z_2}} \right|\]then,
A. \[\operatorname{Im} (\dfrac{{{z_1}}}{{{z_2}}}) = 0\]
B. \[\operatorname{Re} (\dfrac{{{z_1}}}{{{z_2}}}) = 0\]
C. \[\operatorname{Re} (\dfrac{{{z_1}}}{{{z_2}}}) = 0 = \operatorname{Im} (\dfrac{{{z_1}}}{{{z_2}}})\]
D. None of these

Answer 1

Hint:Any term of an equation may be taken from one side to other with the change in its sign, this does not affect the equality of the statement and this process is called transposition. The standard symbol for the set of all complex numbers is $Z = a+ib$, where $a$ and $b$ are real numbers. We will also use the formula \[\cos ({\theta _1} - {\theta _2}) = 1\].

Complete step by step answer:
According to the given information, we have,
\[\left| {{z_1}} \right| = \left| {{z_2}} \right| + \left| {{z_1} - {z_2}} \right|\]
By using transposition in the above equation, we get,
\[ \Rightarrow \left| {{z_1}} \right| - \left| {{z_2}} \right| = \left| {{z_1} - {z_2}} \right|\]
Squaring on both the sides, we get,
\[ \Rightarrow {(\left| {{z_1}} \right| - \left| {{z_2}} \right|)^2} = {\left| {{z_1} - {z_2}} \right|^2}\]
Simplify this above equation, we get,
\[ \Rightarrow {\left| {{z_1}} \right|^2} + {\left| {{z_2}} \right|^2} - 2\left| {{z_1}} \right|\left| {{z_2}} \right| = {\left| {{z_1}} \right|^2} + {\left| {{z_2}} \right|^2} - 2\left| {{z_1}} \right|\left| {{z_2}} \right|\cos ({\theta _1} - {\theta _2})\]

We know that,
\[\cos ({\theta _1} - {\theta _2}) = 1\]
\[ \Rightarrow {\theta _1} - {\theta _2} = 0\]
As we know that, \[\arg ({z_1}) = {\theta _1}\]and \[\arg ({z_2}) = {\theta _2}\]
\[ \Rightarrow \arg ({z_1}) - \arg ({z_2}) = 0\]
\[ \Rightarrow \dfrac{{{z_1}}}{{{z_2}}}\]is purely real, which means that the imaginary part is zero.
\[ \Rightarrow \operatorname{Im} (\dfrac{{{z_1}}}{{{z_2}}}) = 0\]
Thus, if\[{z_1}\]and \[{z_2}\]are two complex numbers that \[\left| {{z_1}} \right| = \left| {{z_2}} \right| + \left| {{z_1} - {z_2}} \right|\]then\[\operatorname{Im} (\dfrac{{{z_1}}}{{{z_2}}}) = 0\].
Hence, option (1) is the correct answer.

Another method to solve this problem is as below:
Let,
\[{z_1} = \cos {\theta _1} + i\sin {\theta _1}\]and\[{z_2} = \cos {\theta _2} + i\sin {\theta _2}\]
\[\therefore {z_1} + {z_2} = \cos {\theta _1} + i\sin {\theta _1} + \cos {\theta _2} + i\sin {\theta _2}\]
\[ \Rightarrow {z_1} + {z_2} = (\cos {\theta _1} + \cos {\theta _2}) + i(\sin {\theta _1} + \sin {\theta _2})\]
Now,
\[\left| {{z_1} + {z_2}} \right| = \left| {{z_1}} \right| + \left| {{z_2}} \right|\]
Substituting the values, we get,
\[ \Rightarrow \sqrt {{{(\cos {\theta _1} + \cos {\theta _2})}^2} + {{(\sin {\theta _1} + \sin {\theta _2})}^2}} = 1 + 1\]

Simplify the above equation, we get,
\[ \Rightarrow \sqrt {{{(\cos {\theta _1} + \cos {\theta _2})}^2} + {{(\sin {\theta _1} + \sin {\theta _2})}^2}} = 2\]
Squaring on both the sides, we get,
\[ \Rightarrow {(\cos {\theta _1} + \cos {\theta _2})^2} + {(\sin {\theta _1} + \sin {\theta _2})^2} = {2^2}\]
\[ \Rightarrow 2(1 + \cos ({\theta _1} - {\theta _2})) = 4\]
Dividing number\[2\]on both the side, we get,
\[ \Rightarrow (1 + \cos ({\theta _1} - {\theta _2})) = \dfrac{4}{2}\]
\[ \Rightarrow 1 + \cos ({\theta _1} - {\theta _2}) = 2\]

By using the transposition in the above equation, we get,
\[ \Rightarrow \cos ({\theta _1} - {\theta _2}) = 2 - 1\]
\[ \Rightarrow \cos ({\theta _1} - {\theta _2}) = 1\]
\[ \Rightarrow {\theta _1} - {\theta _2} = 0\]
As we know that, \[\arg ({z_1}) = {\theta _1}\]and \[\arg ({z_2}) = {\theta _2}\]
\[ \Rightarrow \arg ({z_1}) - \arg ({z_2}) = 0\]
\[ \Rightarrow \dfrac{{{z_1}}}{{{z_2}}}\]is purely real, which means that the imaginary part is zero.
\[ \therefore \operatorname{Im} (\dfrac{{{z_1}}}{{{z_2}}}) = 0\]

Note:The complex number is the combination of a real number and an imaginary number. Either part can be zero. Hence, it is a simple representation of addition of real numbers and an imaginary number in the form of \[a + ib\], where $a$ and $b$ are constants and $i$ is the imaginary part. We also prove this using graphs too.