Question #064a0

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Answer 1 · 2017-04-07T23:28:36

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

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

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

$r = sqrt(1^2+1^2)$

$r = sqrt2$

$theta = tan^-1(b/a)$

$theta = tan^-1(1/1)$

$theta = pi/4" radians"$

$1+i = sqrt2(cos(pi/4)+isin(pi/4))$

Raise both to the sixth power:

$(1+i)^6 = (sqrt(2))^6(cos(6pi/4)+isin(6pi/4))$

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:

$(1+i)^6 = 8(cos(3pi/2)+isin(3pi/2))$

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

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

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