Question #064a0

Question

1278 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-04-07T23:28:36

${(1 + i)}^{6} = 0 - 8 i$ $(1+i)^6 = 0-8i$

Convert, $1 + i$ $1+i$ , from $a + b i$ $a+bi$ to polar form:

$r = \sqrt{a^{2} + b^{2}}$ $r = sqrt(a^2+b^2)$

$r = \sqrt{1^{2} + 1^{2}}$ $r = sqrt(1^2+1^2)$

$r = \sqrt{2}$ $r = sqrt2$

$θ = {tan}^{- 1} (\frac{b}{a})$ $theta = tan^-1(b/a)$

$θ = {tan}^{- 1} (\frac{1}{1})$ $theta = tan^-1(1/1)$

$θ = \frac{π}{4} radians$ $theta = pi/4" radians"$

$1 + i = \sqrt{2} (cos (\frac{π}{4}) + i sin (\frac{π}{4}))$ $1+i = sqrt2(cos(pi/4)+isin(pi/4))$

Raise both to the sixth power:

${(1 + i)}^{6} = {(\sqrt{2})}^{6} (cos (6 \frac{π}{4}) + i sin (6 \frac{π}{4}))$ $(1+i)^6 = (sqrt(2))^6(cos(6pi/4)+isin(6pi/4))$

Please observe that raising the polar form to the 6 power consists of raising the radius to the sixth power and multiplying the angle by 6.

Simplify:

${(1 + i)}^{6} = 8 (cos (3 \frac{π}{2}) + i sin (3 \frac{π}{2}))$ $(1+i)^6 = 8(cos(3pi/2)+isin(3pi/2))$

We can convert the right side back to $a + b i$ $a+bi$ form by substituting the know values for the trigonometric functions:

${(1 + i)}^{6} = 8 (0 + i (- 1))$ $(1+i)^6 = 8(0+i(-1))$

${(1 + i)}^{6} = 0 - 8 i$ $(1+i)^6 = 0-8i$