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Question: Let \({{z}_{1}}\) and \({{z}_{2}}\) be complex numbers such that \({{z}_{1}}\ne {{z}_{2}}\) and \(\l...

Let z1{{z}_{1}} and z2{{z}_{2}} be complex numbers such that z1≠z2{{z}_{1}}\ne {{z}_{2}} and ∣z1∣=∣z2∣\left| {{z}_{1}} \right|=\left| {{z}_{2}} \right| . If Re⁡(z1)>0\operatorname{Re}\left( {{z}_{1}} \right)>0 and Im⁡(z2)<0\operatorname{Im}\left( {{z}_{2}} \right)<0 , then z1+z2z1−z2\dfrac{{{z}_{1}}+{{z}_{2}}}{{{z}_{1}}-{{z}_{2}}} is
a. one
b. real and positive
c. real and negative
d. purely imaginary

Explanation

Solution

Hint: When the modulus of two complex numbers is the same then they are either equal or conjugate of each other. In the given problem they are not equal so they are conjugate. The modulus of two complex numbers may be equal but the numbers may not be equal.

Complete step-by-step answer:

A complex number has two parts, real and imaginary. These represent the x and y coordinates of the point being represented by the complex number. The modulus of a complex number zz is given by ∣z∣=a2+b2\left| z \right|=\sqrt{{{a}^{2}}+{{b}^{2}}} . Here, we have been given with four conditions for such numbers z1≠z2{{z}_{1}}\ne {{z}_{2}},∣z1∣=∣z2∣\left| {{z}_{1}} \right|=\left| {{z}_{2}} \right| , Re⁡(z1)>0\operatorname{Re}\left( {{z}_{1}} \right)>0 and Im⁡(z2)<0\operatorname{Im}\left( {{z}_{2}} \right)<0.This means the number z2{{z}_{2}} is conjugate of number z1{{z}_{1}} .

Let’s assume the numbers as follow,
z1=a+ib............(i){{z}_{1}}=a+ib............(i)
z2=a−ib.......(ii){{z}_{2}}=a-ib.......(ii)

Calculating z1+z2z1−z2\dfrac{{{z}_{1}}+{{z}_{2}}}{{{z}_{1}}-{{z}_{2}}} by substituting the values from equation (i) and equation (ii), we get
z1+z2z1−z2=(a+ib)+(a−ib)(a+ib)−(a−ib) ⇒z1+z2z1−z2=a+ib+a−iba+ib−a+ib ⇒z1+z2z1−z2=2a2ib \begin{aligned} & \dfrac{{{z}_{1}}+{{z}_{2}}}{{{z}_{1}}-{{z}_{2}}}=\dfrac{\left( a+ib \right)+\left( a-ib \right)}{\left( a+ib \right)-\left( a-ib \right)} \\\ & \Rightarrow \dfrac{{{z}_{1}}+{{z}_{2}}}{{{z}_{1}}-{{z}_{2}}}=\dfrac{a+ib+a-ib}{a+ib-a+ib} \\\ & \Rightarrow \dfrac{{{z}_{1}}+{{z}_{2}}}{{{z}_{1}}-{{z}_{2}}}=\dfrac{2a}{2ib} \\\ \end{aligned}

Above equation can be represented as,
⇒z1+z2z1−z2=1×aib...............(iii)\Rightarrow \dfrac{{{z}_{1}}+{{z}_{2}}}{{{z}_{1}}-{{z}_{2}}}=\dfrac{1\times a}{ib}...............(iii)
As, we know i2=−1{{i}^{2}}=-1 then, 1=(−1)2=(i2)2=ι41={{\left( -1 \right)}^{2}}={{\left( {{i}^{2}} \right)}^{2}}={{\iota }^{4}}

Substituting the values in equation (iii), we get

& \Rightarrow \dfrac{{{z}_{1}}+{{z}_{2}}}{{{z}_{1}}-{{z}_{2}}}=\dfrac{{{i}^{4}}\times a}{ib} \\\ & \Rightarrow \dfrac{{{z}_{1}}+{{z}_{2}}}{{{z}_{1}}-{{z}_{2}}}=\dfrac{{{i}^{3}}\times a}{b} \\\ & \Rightarrow \dfrac{{{z}_{1}}+{{z}_{2}}}{{{z}_{1}}-{{z}_{2}}}=\dfrac{i({{i}^{2}})\times a}{b} \\\ & \Rightarrow \dfrac{{{z}_{1}}+{{z}_{2}}}{{{z}_{1}}-{{z}_{2}}}=\dfrac{-a}{b}i \\\ \end{aligned}$$ Hence, the number we got is negative and purely imaginary. Therefore, $\dfrac{{{z}_{1}}+{{z}_{2}}}{{{z}_{1}}-{{z}_{2}}}$ is negative and purely imaginary number. This means the final answer is option (d). Note: The chances of mistakes are if the conjugate of number is not taken correctly or there may be mistakes during the interpretation of the conditions given in the question. One should not get confused with the modulus and the number.