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How solve this equation in complex? 1+z+z^2+...+z^n*z=0

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rouche | Student, Undergraduate | (Level 1) Honors

Posted May 6, 2013 at 4:10 PM via web

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How solve this equation in complex?

1+z+z^2+...+z^n*z=0

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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted May 6, 2013 at 5:34 PM (Answer #3)

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You need to solve for complex z the given equation, such that:

`1 + z + z^2 + z^3 + ... + z^(n+1) = 0`

Multiplying by (z - 1) both sides yields:

`(1 + z + z^2 + z^3 + ... + z^(n+1))(z - 1) = 0*(z - 1)`

You need to convert the left side product into the following binomial difference, such that:

`z^(n + 1) - 1 = 0 => z^(n + 1) = 1`

You need to convert the algebraic form of complex number 1 into trigonometric form, such that:

`z^(n + 1) = cos 0^o + i*sin 0^o`

`z = (cos 0^o + i*sin 0^o)^(1/(n+1))`

Using De Moivre's theorem yields:

`z = (cos ((0^o + 2*k*pi)/(n+1)) + i*sin((0^o + 2*k*pi)/(n+1)))`

You need to give integer values to k, from 0 to n, such that:

`k = 0 => z_1 = cos 0^o + i*sin 0^o = 1`

`k = 1 => z_2 = (cos ((2pi)/(n+1)) + i*sin((2pi)/(n+1)))`

....

`k = n => z_(n+1) = (cos ((2npi)/(n+1)) + i*sin((2npi)/(n+1)))`

Hence, evaluating the n roots of the given equation, using De Moivre's theorem, yields `z_(1,2,...,n+1) = (cos ((0^o + 2*k*pi)/(n+1)) + i*sin((0^o + 2*k*pi)/(n+1))).`

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user8121921 | Student, Grade 10 | (Level 1) eNoter

Posted May 6, 2013 at 4:27 PM (Answer #1)

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If z=1 then the left side <> 0 => z<>1

1+z+z^2+...+z^n*z=[z^(n*z+1)-1]/[z-1]=0

<=> z^(n*z+1)-1=0

<=> z^(n*z+1)=1

if (n*z+1) mod 2 = 1 then this equation has no solution

else z=-1

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user8121921 | Student, Grade 10 | (Level 1) eNoter

Posted May 6, 2013 at 4:57 PM (Answer #2)

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If z=1 then the left side <> 0 => z<>1

1+z+z^2+...+z^n*z=[z^(n*z+1)-1]/[z-1]=0

<=> z^(n*z+1)-1=0

<=> z^(n*z+1)=1

<=> z=1 and z=-1(if n*z mod 2=1)

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