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


4 years ago

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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• Test paper with Video Solution
• Mind Map
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