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Given modulus z = 1, show how solve inequality?  modulus(z/conjugate z + conjugate...

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yapayapa | (Level 2) Honors

Posted June 2, 2013 at 1:52 PM via web

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Given modulus z = 1, show how solve inequality? 

modulus(z/conjugate z + conjugate z/z) < or equal 1

 

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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted June 2, 2013 at 2:29 PM (Answer #1)

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You need to solve the inequality `|z/(bar z) + (bar z)/z| <= 1` , using the provided information that `|z| = 1` , such that:

`|z/(bar z) + (bar z)/z| <= 1 => |(z^2 + (bar z)^2)/(z*(bar z))| <= 1`

You should notice that bar `z = x - i*y` represents the conjugate of complex number `z = x + i*y.`

You may use the polar form of complex numbers `z,bar z` , such that:

`z = cos alpha + i*sin alpha => z^2 = cos^2 alpha + 2i*sin alpha*cos alpha + sin^2 alpha`

Since `i^2 = -1` yields:

`z^2 =cos^2 alpha - sin^2 alpha + 2i*sin alpha*cos alpha`

Replacing `cos 2 alpha` for `cos^2 alpha - sin^2 alpha` yields:

`z^2 = cos 2 alpha + 2i*sin alpha*cos alpha`

`bar z = cos alpha - i*sin alpha => (bar z)^2 = cos^2 alpha - 2i*sin alpha*cos alpha + sin^2 alpha`

`(bar z)^2 = cos^2 alpha - 2i*sin alpha*cos alpha + i^2*sin^2 alpha`

Since `i^2 = -1` yields:

`(bar z)^2 =cos^2 alpha -sin^2 alpha - 2i*sin alpha*cos alpha`

Replacing `cos 2 alpha` for `cos^2 alpha -sin^2 alpha` yields:

`(bar z)^2 = cos 2 alpha - 2i*sin alpha*cos alpha`

`z*bar z = (cos alpha + i*sin alpha)(cos alpha - i*sin alpha)`

`z*bar z = cos^2 alpha - i^2*sin^2 alpha`

Since `i^2 = -1` yields:

`z*bar z = cos^2 alpha + sin^2 alpha = 1`

Replacing `1 + 2i*sin alpha*cos alpha` for `z^2, 1 - 2i*sin alpha*cos alpha` for `(bar z)^2` and 1 for `z*bar z` yields:

`|(cos 2 alpha + 2i*sin alpha*cos alpha + cos 2 alpha - 2i*sin alpha*cos alpha)/1| <= 1`

Reducing duplicate members yields:

`|2cos 2alpha| <= 1 `

Using the definition of absolute value yields:

`-1 <= 2cos 2 alpha <= 1 => -1/2 <= cos 2alpha <= 1/2 => {(pi/3 <= 2alpha <= (2pi)/3),((4pi)/3 <= 2alpha <= (5pi)/3):} => alpha in [pi/6,pi/3]U[(2pi)/3,(5pi)/6]`

Hence, evaluating the values of alpha in `z = cos alpha + i*sin alpha` , that make the given inequality valid, yields `alpha in [pi/6,pi/3]U[(2pi)/3,(5pi)/6].`

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