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Grade: 12th pass 

                        
The minimum |z| satisfying |z+1/z| = 2 is ...........?

							
The minimum |z| satisfying |z+1/z| = 2 is ...........?
4 years ago

Answers : (3)

Vikas TU
14149 Points 
							
Put z = x+iy
=> |x + iy + 1/(x + iy)|
=> |[(x+iy)^2 + 1]/(x+iy)|
=>|x^2 – y^2 + 2xyi + 1|/|x+iy|
=> root[(x^2 – y^2 +1)^2 + 1]/root(x^2 + y^2)
Now solve it and compare it.
4 years ago
Anshuman Mohanty
27 Points 
							
Please read the question again. I want to find the minimum |z| 
Please tell me how to use Triangle Inequality in this..
4 years ago
Ajay
209 Points 
							
The answer is -1+\small \sqrt{2} as  shown below......................................................................................
\left| z+\frac { 1 }{ z } \right| \quad =\quad 2\\ \left| z+\frac { \bar { z } }{ { |z| }^{ 2 } } \right| \quad =\quad 2\\ \\ now\quad \left| |z|-\left| \frac { \bar { z } }{ { |z| }^{ 2 } } \right| \right| <=\left| z+\frac { \bar { z } }{ { |z| }^{ 2 } } \right| \\ \quad \quad \quad \quad \left| |z|-\frac { |z| }{ { |z| }^{ 2 } } \right| <=2\\ \quad \quad \quad \quad \quad \quad \left| |z|-\frac { 1 }{ { |z| } } \right| <=2\\ -2\le |z|-\frac { 1 }{ { |z| } } <2\\ taking\quad first\quad inequality\quad -2\le |z|-\frac { 1 }{ { |z| } } \quad will\quad give\quad the\quad minimum\quad value\\ This\quad becomes\quad { |z| }^{ 2 }+2|z|-1\quad \ge \quad 0\\ (|z|-(-1+\sqrt { 2 } )(|z|-(-1-\sqrt { 2 } )\quad \ge \quad 0\\ \\ |z|\ge (-1+\sqrt { 2 } )\\ \\
 
 
 
4 years ago
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