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Calculate in C ecuation: (1-i)z^6=1-i√3

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alexa0048 | Student, Grade 11 | (Level 3) eNoter

Posted March 25, 2013 at 4:06 PM via web

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Calculate in C ecuation:

(1-i)z^6=1-i√3

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pramodpandey | College Teacher | (Level 3) Valedictorian

Posted March 27, 2013 at 7:35 AM (Answer #1)

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`(1-i)Z^6=1-isqrt(3)`

`Z^6=(1-isqrt(3))/(1-i)`

`Z^6=((1-isqrt(3))(1+i))/{(1-i)(1+i)}`

`Z^6=(1-isqrt(3)+i-i^2sqrt(3))/(1-i^2)`

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

`(1+sqrt(3)+i(1-sqrt(3))/2)=(1+sqrt(3))/2+i(1-sqrt(3))/2 `

Let change `(1+sqrt(3))/2+i(1-sqrt(3))/2 ` in ploar coordinate system.

`(1+sqrt(3))/2+i(1-sqrt(3))/2=rcos(theta)+irsin(theta)`

`rcos(theta)=(1+sqrt(3))/2`

`rsin(theta)=(1-sqrt(3))/3`

squaring and adding ,we get

`r^2=2`

`r=sqrt(2)`

also  if we divide ,and simplify ,we get

`tan(theta)=(1-sqrt(3))/(1+sqrt(3))`

`tan(theta)=tan(-pi/12)`

`` Thus  we have now,

`Z^6=sqrt(2)(cos(theta)+isin(theta))`

`Z={sqrt(2)(cos(theta)+isin(theta))}^(1/6)`

`Z=2^(1/12)(cos((theta+2kpi)/6)+isin((theta+2kpi)/6))`

`k=0,1,2,3,4,5 and theta=-pi/12`

 

`` Ans.

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