` ` calculated : (1-i)(z*z*z*z*z*z)=1-isqrt(3)

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to find the solutions to the equation `(1 - i)*z^6 = 1 - sqrt3*i` , hence, you need to isolate `z^6` to the left side, such that:

`z^6 = (1 - sqrt3*i)/(1 - i)`

You need to multiplicate by the conjugate of `1 - i` , such that:

`z^6 = ((1 - sqrt3*i)(1 + i))/((1 - i)(1 + i))`

`z^6 = ((1 - sqrt3*i)(1 + i))/(1 - i^2)`

Replacing -1 for `i^2` yields:

`z^6 = ((1 - sqrt3*i)(1 + i))/2`

`z^6 = (1 + i - sqrt 3*i - sqrt 3*i^2)/2 `

`z^6 = (1 + sqrt3 + i*(1 - sqrt 3))/2 `

You may find the solutions to the given equation, using De Moivre's theorem, hence, you need to convert the rectangular form of complex number `(1 + sqrt3 + i*(1 - sqrt 3))/2` into polar form, such that:

`z^6 = |z|*(cos alpha + i*sin alpha)`

`|z| = sqrt((1 + sqrt3)^2 + (1 - sqrt3)^2)`

`|z| = sqrt(1 + 2sqrt3 + 3 + 1 - 2sqrt3 + 3)`

`|z| = 2sqrt2`

`tan alpha = (1 - sqrt3)/(1 + sqrt3) => tan alpha = (1 - sqrt3)^2/(1 - 3)`

`tan alpha = (1 - 2sqrt3 + 3)/(-2)`

`tan alpha = -2+ sqrt 3 => alpha = tan^(-1)(-2+ sqrt 3)`

`z^6 = 2sqrt2*(cos (tan^(-1)(-2+ sqrt 3)) + i*sin (tan^(-1)(-2+ sqrt 3)))`

`z = (2sqrt2)^(1/6)(cos (tan^(-1)(-2+ sqrt 3)) + i*sin (tan^(-1)(-2+ sqrt 3)))^(1/6)`

Using De Moivre's theorem yields:

`z = 2^(1/4)(cos (tan^(-1)(-2+ sqrt 3) + 2npi)/6 + i*sin (tan^(-1)(-2+ sqrt 3) + 2npi)/6)`

You need to consider `n=0,1,2,3,4,5` to evaluate the solutions to the given equation.

Hence, evaluating the solutions to the given equation, using De Moivre's theorem, yields `z = 2^(1/4)(cos (tan^(-1)(-2+ sqrt 3) + 2npi)/6 + i*sin (tan^(-1)(-2+ sqrt 3) + 2npi)/6), n=0,1,2,3,4,5.`

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