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

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

Anshuman Mohanty , 8 Years ago

Grade 12th pass

3 Answers

Vikas TU

Last Activity: 8 Years ago

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.

Anshuman Mohanty

Last Activity: 8 Years ago

Please read the question again. I want to find the minimum |z|
Please tell me how to use Triangle Inequality in this..

Ajay

Last Activity: 8 Years ago

The answer is -1+ $\small \sqrt{2}$ as shown below......................................................................................

$\left| z+\frac { 1 }{ z } \right| \quad =\quad 2\\ \left| z+\frac { \bar { z } }{ { |z| }^{ 2 } } \right| \quad =\quad 2\\ \\ now\quad \left| |z|-\left| \frac { \bar { z } }{ { |z| }^{ 2 } } \right| \right| <=\left| z+\frac { \bar { z } }{ { |z| }^{ 2 } } \right| \\ \quad \quad \quad \quad \left| |z|-\frac { |z| }{ { |z| }^{ 2 } } \right| <=2\\ \quad \quad \quad \quad \quad \quad \left| |z|-\frac { 1 }{ { |z| } } \right| <=2\\ -2\le |z|-\frac { 1 }{ { |z| } } <2\\ taking\quad first\quad inequality\quad -2\le |z|-\frac { 1 }{ { |z| } } \quad will\quad give\quad the\quad minimum\quad value\\ This\quad becomes\quad { |z| }^{ 2 }+2|z|-1\quad \ge \quad 0\\ (|z|-(-1+\sqrt { 2 } )(|z|-(-1-\sqrt { 2 } )\quad \ge \quad 0\\ \\ |z|\ge (-1+\sqrt { 2 } )\\ \\$