Question

If z is a complex number such that $\left| z \right|\ge 2$ then the minimum value of $\left| z+\dfrac{1}{2} \right|$ is

& \text{A}.\text{ is equal to }\dfrac{5}{2} \\\ & \text{B}.\text{ lies in interval }\left( \text{1},\text{ 2} \right) \\\ & \text{C}.\text{ is strictly greater than }\dfrac{5}{2} \\\ & \text{D}.\text{ is strictly greater than }\dfrac{3}{2}\text{ but less than }\dfrac{5}{2} \\\ \end{aligned}$$

Answer 1

To solve this question, we will use formula of addition and subtraction of two complex number which are given as:
$\left| {{z}_{1}}+{{z}_{2}} \right|\le \left| {{z}_{1}} \right|+\left| {{z}_{2}} \right|\text{ and }\left| {{z}_{1}}-{{z}_{2}} \right|\ge \left| {{z}_{1}} \right|-\left| {{z}_{2}} \right|$
Finally, we will substitute ${{z}_{1}}=z\text{ and }{{\text{z}}_{\text{2}}}=-\dfrac{1}{2}$ to get maximum and minimum value of $\left| z+\dfrac{1}{2} \right|$

Complete step-by-step answer:
Given that, $\left| z \right|\ge 2$
We have two formulas of add and subtraction of two complex number which are given as:
$\left| {{z}_{1}}+{{z}_{2}} \right|\le \left| {{z}_{1}} \right|+\left| {{z}_{2}} \right|\text{ and }\left| {{z}_{1}}-{{z}_{2}} \right|\ge \left| {{z}_{1}} \right|-\left| {{z}_{2}} \right|$
Let us assume the value ${{z}_{1}}=z$ and value of ${{\text{z}}_{\text{2}}}=-\dfrac{1}{2}$ as we have to calculate the value of $\left| z+\dfrac{1}{2} \right|$
Let $\left| {{z}_{1}}-{{z}_{2}} \right|\ge \left| {{z}_{1}} \right|-\left| {{z}_{2}} \right|\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. (i)}$
Using equation (i) by putting ${{z}_{1}}=z\text{ and }{{\text{z}}_{\text{2}}}=-\dfrac{1}{2}$ we get:
$\left| z-\left( \dfrac{-1}{2} \right) \right|\text{ }\ge \text{ }\left| \text{z} \right|-\left| -\dfrac{1}{2} \right|$
Now, as value of $\left| z \right|\ge 2$
$\Rightarrow \left| z \right|-\left| \dfrac{-1}{2} \right|\ge 2-\left| \dfrac{-1}{2} \right|$
Also, as any value under mode comes out to be positive. So, $\left| \dfrac{-1}{2} \right|=+\dfrac{1}{2}$
Using this in above, we get:

& \Rightarrow \left| \left( z \right)-\left( \dfrac{-1}{2} \right) \right|\ge 2-\dfrac{1}{2} \\\ & \Rightarrow \left| \left( z \right)-\left( \dfrac{-1}{2} \right) \right|\ge \dfrac{2-1}{2} \\\ \end{aligned}$$ Taking LCM on right hand side $$\begin{aligned} & \Rightarrow \left| z+\dfrac{1}{2} \right|\ge \dfrac{4-1}{2} \\\ & \Rightarrow \left| z+\dfrac{1}{2} \right|\ge \dfrac{3}{2} \\\ \end{aligned}$$ So, we get that the value of $$\left| z+\dfrac{1}{2} \right|\ge \dfrac{3}{2}$$ consider $$\left| {{z}_{1}}+{{z}_{2}} \right|\le \left| {{z}_{1}} \right|+\left| {{z}_{2}} \right|$$ Let ${{z}_{1}}=z\text{ and }{{\text{z}}_{\text{2}}}=\dfrac{1}{2}$ in above, we get: $$\left| z+\dfrac{1}{2} \right|\le \left| z \right|+\left| \dfrac{1}{2} \right|$$ Now, value of $\left| z \right|\ge 2$ $$\begin{aligned} & \Rightarrow \left| z+\dfrac{1}{2} \right|\le 2+\dfrac{1}{2} \\\ & \Rightarrow \left| z+\dfrac{1}{2} \right|\le \dfrac{5}{2} \\\ \end{aligned}$$ So, the value of $\left| z+\dfrac{1}{2} \right|$ is greater than $\dfrac{3}{2}$ and less than $\dfrac{5}{2}$ **So, the correct answer is “Option D”.** **Note:** Another way to get that $\left| {{z}_{2}} \right|=+\text{ positive}$ is, let $$z=x+y\text{ and }\left| z \right|=\sqrt{{{x}^{2}}+{{y}^{2}}}\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. (ii)}$$ Now, we have $$\begin{aligned} & {{z}_{2}}=-\dfrac{1}{2}\Rightarrow {{z}_{2}}=-\dfrac{1}{2}+0 \\\ & \Rightarrow \left| {{z}_{2}} \right|=\sqrt{{{\left( -\dfrac{1}{2} \right)}^{2}}+{{\left( 0 \right)}^{2}}} \\\ & \Rightarrow \left| {{z}_{2}} \right|=\dfrac{1}{2} \\\ \end{aligned}$$ This above is obtained by using equation (ii) so, we can easily use $\left| {{z}_{2}} \right|=\dfrac{1}{2}$ to get the result. Hence, by this way we can avoid the confusion part that ${{z}_{2}}$ can be negative.