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#