Use DeMoivres theorem to find `(-2i)^6` .

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You should remember De Moivre's identity used for powers of complex number z = cos alpha + i*sin alpha such that:

z^n = r^n(cos alpha + i*sin alpha)^n = r^n(cos (n*alpha) + i*sin (n*alpha))

Since De Moivre's identity works for trigonometric form of complex numbers, you need to transform the given rectangular form of the complex number z = -2i into the trigonometric form, such that:

r = sqrt(a^2 + b^2)

Identifying the real part and the imaginary parts of the given complex number, yields:

a = 0, b = -2

r = sqrt(0^2 + (-2)^2) => r = 2

tan alpha = b/a => alpha = arctan(b/a) => alpha = arctan(-2/0) =>alpha = -pi/2 => alpha = 3pi/2

Hence, evaluating the trigonometric form of teh complex number `z = -2i` yields:

`z = 2(cos((3pi)/2) + i*sin((3pi)/2))`

You need to raise to the 6th power, such that:

`z^6 = (2(cos((3pi)/2) + i*sin((3pi)/2)))^6`

Using De Moivre's identity, yields:

`z^6 = (2^6)(cos((3pi*6)/2) + i*sin((3pi*6)/2))`

`z^6 = 64*(cos(9pi) + i*sin(9pi))`

`sin(9pi) = sin(8pi + pi) = sin 8pi cos pi + sin pi cos 8pi`

Since `sin pi = 0` and `cos pi = -1` yields:

`sin(9pi) = -sin 8pi`

`sin (8pi) = sin(6pi + 2pi) =sin 6pi cos 2pi + sin 2pi cos 6pi`

`sin (8pi) = sin 6pi = sin 4pi = sin 2pi = 0 => sin(9pi) = 0`

Evaluating `cos 9pi` yields:

`cos 9pi = cos(8pi + pi) = cos 8pi cos pi - sin 8pi sin pi`

Since `sin 8pi = sin pi = 0` and `cos pi = -1` yields:

`cos 9pi = -cos 8pi`

`cos 8pi = cos(6pi + 2pi) = cos 6pi cos 2pi - sin 6pi sin 2pi`

`cos 8pi = cos 6pi = cos 4pi = cos 2pi = 1`

`cos 9pi = -cos 8pi => cos 9pi = -1`

Hence, substituting -1 for `cos 9 pi` and 0 for `sin 9pi` yields:

`z^6 = 64*(-1 + i*0) => z^6 = -64`

Hence, evaluating `z^6` , using De Moivre's identity yields `z^6 = -64.`

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lemjay | High School Teacher | (Level 3) Senior Educator

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To find `(x+yi)^n` using De Moivre's Theorem, we need to convert `x+yi` to polar form `r(costheta + isintheta)` first. 

To determine r and `theta` , use the formulas:

`r = sqrt(x^2+y^2)`          and          ` theta = tan^(-1)y/x`

Re-writing -2i as 0+2i, its r and `theta` is:

`r=sqrt(0^2+(-2)^2)=sqrt4=2 `               

Since x=0 and y is positive, then `theta=90` (based on a unit circle chart).

Then, substitute r and `theta` to the polar formula.

`-2i = 2 (cos 90 + isin 90)`

And, replace -2i with `2(cos90+isin90)` in `(-2i)^6` . So,

`(-2i)^6 = (2(cos90+isin90))^6`

From here, apply De Moivre's theorem which is:

`(r(costheta+isintheta))^n= r^n(cos(ntheta)+isin(ntheta))`

So,

`(2(cos90+isin90))^6 = 2^6 (cos(6*90) + isin(6*90))`

`(2(cos90+isin90))^6 =64(cos 540 + isin 540)`

`(2(cos90+isin90))^6 = 64(-1+i*0)`

`(2(cos90+isin90))^6=64(-1)`

`(2(cos90+isin90))^6=-64`

Hence, `(-2i)^6 =-64` .

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