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we have the modulus of a complex numbe as \[\left| {{z}_{1}} \right|=2\] and \[\left( 1-i \right){{z}_{2}}+\left( 1+i \right)\overline{{{z}_{2}}}=8\sqrt{2}\] then the minimum value of \[\left| {{z}_{1}}-{{z}_{2}} \right|\] is: -
(a) 2
(b) 4
(c) 1
(d) \[\sqrt{2}\]

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Last updated date: 20th Jun 2024
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Answer
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Hint: Assume \[{{z}_{1}}={{x}_{1}}+i{{y}_{1}}\] and \[{{z}_{2}}={{x}_{2}}+i{{y}_{2}}\]. Find the conjugate of \[{{z}_{2}}\] by replacing ‘+’ sign with ‘-’ sign in the expression of \[{{z}_{2}}\]. Now, find the relation between \[{{x}_{1}}\] and \[{{y}_{1}}\] to trace the curve on which \[{{z}_{1}}\] lies. Similarly, find the relation between \[{{x}_{2}}\] and \[{{y}_{2}}\] to trace the curve on which \[{{z}_{2}}\] lies. Use the formula: - \[\left| z \right|=\sqrt{{{x}^{2}}+{{y}^{2}}}\] for \[{{z}_{1}}\]. Finally, find the minimum value of \[\left| {{z}_{1}}-{{z}_{2}} \right|\] by finding the distance between the closest points on the two curves.

Complete step-by-step solution
We have been given, \[\left| {{z}_{1}} \right|=2\]. Let us assume, \[{{z}_{1}}={{x}_{1}}+i{{y}_{1}}\].
We know that, \[\left| z \right|=\sqrt{{{x}^{2}}+{{y}^{2}}}\],
\[\begin{align}
  & \Rightarrow \left| {{z}_{1}} \right|=2 \\
 & \Rightarrow \sqrt{{{x}_{1}}^{2}+{{y}_{1}}^{2}}=2 \\
\end{align}\]
\[\Rightarrow {{x}_{1}}^{2}+{{y}_{1}}^{2}={{2}^{2}}\], this is the equation of a circle.
Hence, \[{{z}_{1}}\] lies on a circle with a radius of 2 units.
Now, we have: - \[\left( 1-i \right){{z}_{2}}+\left( 1+i \right)\overline{{{z}_{2}}}=8\sqrt{2}\]. Assuming, \[{{z}_{2}}={{x}_{2}}+i{{y}_{2}}\], we get,
\[{{z}_{2}}={{x}_{2}}+i{{y}_{2}}\]
\[\Rightarrow \overline{{{z}_{2}}}\] = conjugate of \[{{z}_{2}}\] = \[{{x}_{2}}-i{{y}_{2}}\].
So, the expression becomes: -
\[\begin{align}
  & \Rightarrow \left( 1-i \right)\left( {{x}_{2}}+i{{y}_{2}} \right)+\left( 1+i \right)\left( {{x}_{2}}-i{{y}_{2}} \right)=8\sqrt{2} \\
 & \Rightarrow \left( {{x}_{2}}-i{{x}_{2}}+i{{y}_{2}}-{{i}^{2}}{{y}_{2}} \right)+\left( {{x}_{2}}+i{{x}_{2}}-i{{y}_{2}}-{{i}^{2}}{{y}_{2}} \right)=8\sqrt{2} \\
\end{align}\]
Substituting, \[{{i}^{2}}=-1\] and simplifying we get,
\[\Rightarrow 2{{x}_{2}}+2{{y}_{2}}=8\sqrt{2}\]
\[\Rightarrow {{x}_{2}}+{{y}_{2}}=4\sqrt{2}\], this is the equation of a straight line.
Hence, \[{{z}_{2}}\] lies on a straight line.
 
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Clearly, we can see that A and B are the closest points.
So, the minimum distance between the curves is AB.
\[\Rightarrow {{\left| {{z}_{1}}-{{z}_{2}} \right|}_{\min }}=AB\]
\[\Rightarrow {{\left| {{z}_{1}}-{{z}_{2}} \right|}_{\min }}=OA-OB\] - (i)
Here, OB is the radius of the circle.
\[\Rightarrow OB=2\]units.
Now, OA is the minimum distance of the line from the origin O and we know that the shortest path of a line from a point is the perpendicular distance between them. So, OA is perpendicular to line \[x+y=4\sqrt{2}\].
Applying the formula to find the shortest path between a point \[\left( \alpha ,\beta \right)\] and the line \[ax+by+c=0\], we have,
\[\Rightarrow d=\left| \dfrac{a\alpha +b\beta +c}{\sqrt{{{a}^{2}}+{{b}^{2}}}} \right|\], d = minimum distance.
So, applying this formula for the point (0, 0) and line, \[x+y-4\sqrt{2}=0\], we get,
\[\begin{align}
  & \Rightarrow OA=\left| \dfrac{0\times 1+0\times 1-4\sqrt{2}}{\sqrt{{{1}^{2}}+{{1}^{2}}}} \right| \\
 & \Rightarrow OA=\left| \dfrac{-4\sqrt{2}}{\sqrt{2}} \right| \\
\end{align}\]
\[\Rightarrow \] OA = 4 units.
Substituting OA = 4 and OB = 2 in relation (i), we get,
\[\Rightarrow {{\left| {{z}_{1}}-{{z}_{2}} \right|}_{\min }}=4-2=2\]
Hence, option (a) is the correct answer.

Note: One may note that it is very important to find the relation between the variables x and y to trace the curve and find the minimum of the expression, \[\left| {{z}_{1}}-{{z}_{2}} \right|\]. If will not apply the above process then it will be very difficult to solve the problem. Also, remember that tracing the figure of the curve is very important to see the minimum distance of the points otherwise we can get confused.