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Question: Let \({{z}_{1}}\text{ and }{{z}_{2}}\) be two complex numbers satisfying \(\left| {{z}_{1}} \right|=...

Let z1 and z2{{z}_{1}}\text{ and }{{z}_{2}} be two complex numbers satisfying ∣z1∣=9 and ∣z2−3−4i∣=4\left| {{z}_{1}} \right|=9\text{ and }\left| {{z}_{2}}-3-4i \right|=4 then, the minimum value of ∣z1−z2∣\left| {{z}_{1}}-{{z}_{2}} \right| is

& \text{A}.\text{ }0 \\\ & \text{B}.\text{ 1} \\\ & \text{C}.\text{ }\sqrt{2} \\\ & \text{D}.\text{ 2} \\\ \end{aligned}$$
Explanation

Solution

To solve this question we will first of all use a property of complex number stated below:
∣z1−z2∣≥∣z1∣−∣z2∣\left| {{z}_{1}}-{{z}_{2}} \right|\ge \left| {{z}_{1}} \right|-\left| {{z}_{2}} \right|
We will use the value ∣z2−3−4i∣=4\left| {{z}_{2}}-3-4i \right|=4 in above formula to calculate ∣z2∣\left| {{z}_{2}} \right| or least or maximum value of ∣z2∣\left| {{z}_{2}} \right|. In between this we will use that, if z=a+ibz=a+ib then mode of ∣z∣=a2+b2\left| z \right|=\sqrt{{{a}^{2}}+{{b}^{2}}}. After calculating maximum value of ∣z2∣\left| {{z}_{2}} \right| we will again apply ∣z1−z2∣≥∣z1∣−∣z2∣\left| {{z}_{1}}-{{z}_{2}} \right|\ge \left| {{z}_{1}} \right|-\left| {{z}_{2}} \right| to calculate minimum value of ∣z1−z2∣\left| {{z}_{1}}-{{z}_{2}} \right|

Complete step-by-step answer:
We have a property of complex number stated as below:
If Z1 and Z2{{Z}_{1}}\text{ and }{{Z}_{2}} are two complex numbers then ∣Z1−Z2∣≥∣Z1∣−∣Z2∣\left| {{Z}_{1}}-{{Z}_{2}} \right|\ge \left| {{Z}_{1}} \right|-\left| {{Z}_{2}} \right|

We are given z1 and z2{{z}_{1}}\text{ and }{{z}_{2}} two complex numbers and ∣z1∣=9 and ∣z2−3−4i∣=4\left| {{z}_{1}} \right|=9\text{ and }\left| {{z}_{2}}-3-4i \right|=4
Consider ∣z2−3−4i∣\left| {{z}_{2}}-3-4i \right|
Taking minus common, we get:
∣z2−(3+4i)∣\left| {{z}_{2}}-\left( 3+4i \right) \right|
Using formula stated above by using Z2=(3+4i){{Z}_{2}}=\left( 3+4i \right) we get:
∣z2−(3+4i)∣≥∣z2∣−∣3+4i∣ . . . . . . . . . . . . (i)\left| {{z}_{2}}-\left( 3+4i \right) \right|\ge \left| {{z}_{2}} \right|-\left| 3+4i \right|\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. (i)}
We have a complex number z=a+ibz=a+ib then the mode of z is ∣z∣=a2+b2\left| z \right|=\sqrt{{{a}^{2}}+{{b}^{2}}}
We have ∣3+4i∣\left| 3+4i \right| applying formula stated above to calculate ∣3+4i∣\left| 3+4i \right| we get:

& \left| 3+4i \right|=\sqrt{{{\left( 3 \right)}^{2}}+{{\left( 4 \right)}^{2}}} \\\ & \left| 3+4i \right|=\sqrt{9+16} \\\ & \left| 3+4i \right|=\sqrt{25} \\\ & \left| 3+4i \right|=\pm 5 \\\ \end{aligned}$$ Now, as the LHS of the above equation has mode = negative value is not possible. $$\Rightarrow \left| 3+4i \right|=\pm 5$$ Using this in equation (i), we get: $$\left| {{z}_{2}}-\left( 3+4i \right) \right|\ge \left| {{z}_{2}} \right|-5$$ Now, as $$\begin{aligned} & \left| {{z}_{2}}-\left( 3+4i \right) \right|=4 \\\ & \Rightarrow 4\ge \left| {{z}_{2}} \right|-5 \\\ \end{aligned}$$ Adding 5 both sides, we get: $$\begin{aligned} & 4+5\ge \left| {{z}_{2}} \right| \\\ & \Rightarrow \left| {{z}_{2}} \right|\le 9\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. (ii)} \\\ \end{aligned}$$ Now, we have for two complex number ${{z}_{1}}\text{ and }{{z}_{2}}$ $$\left| {{z}_{1}}-{{z}_{2}} \right|\ge \left| {{z}_{1}} \right|-\left| {{z}_{2}} \right|$$ Then, minimum value of $$\left| {{z}_{1}}-{{z}_{2}} \right|=\left| {{z}_{1}} \right|-{{\left| {{z}_{2}} \right|}_{\max }}$$ Minimum value of $$\left| {{z}_{1}}-{{z}_{2}} \right|=\left| {{z}_{1}} \right|-{{\left| {{z}_{2}} \right|}_{\max }}$$ from equation (ii) we have ${{\left| {{z}_{2}} \right|}_{\max }}=9$ and given in question is $\left| {{z}_{1}} \right|=9$ Minimum value of $$\Rightarrow \left| {{z}_{1}}-{{z}_{2}} \right|=9-9=0$$ Therefore, the minimum value of $$\left| {{z}_{1}}-{{z}_{2}} \right|=0$$ so **So, the correct answer is “Option A”.** **Note:** If a number is of the form $\left| z \right|\le a$ then minimum value of $\left| z \right|$ is not known but maximum value of $\left| z \right|=a$ this is what we have used for $\max \left| {{z}_{2}} \right|\Rightarrow \left| {{z}_{2}} \right|\le 9$ So, $\max \left( \left| {{z}_{2}} \right| \right)=9$ A key point to note is, at the end of solution where we have used $$\min \left| {{z}_{1}}-{{z}_{2}} \right|=\left| {{z}_{1}} \right|-{{\left| {{z}_{2}} \right|}_{\max }}$$ This is so as if we have a difference of two numbers x-y=a then as larger the value of y, we get the least value of a because subtracting larger value gives least a. Therefore, the step $$\min \left| {{z}_{1}}-{{z}_{2}} \right|=\left| {{z}_{1}} \right|-{{\left| {{z}_{2}} \right|}_{\max }}$$ is correct.