Using DeMoivre;s theorem compute the complex number #z= ( ( i -3)/(2+i))^7# ?

1 Answer
Hriman
Jul 5, 2018

#z=-8-8i# in standard form

#z= 8sqrt2(cos((5pi)/4)+isin((5pi)/4))# in trigonometric form

Explanation:

Before we use DeMoivre's Theorem, let's divide:

#((i-3)/(2+i)*(2-i)/(2-i))^7=((5i-5)/5)^7=(-1+i)^7#

Now let's convert to trigonometric form so we can apply the theorem:

#r=sqrt(x^2+y^2)#
#r=sqrt((-1)^2+1^2)#
#r=sqrt2#

#theta=arctan(y/x)#
With the angles, always note the quadrant you're in, in this case with a negative x-value and a positive y-value, you will be in the 2nd quadrant.
#theta=arctan((1)/-1)=-45^@ +180^@=135^@=(3pi)/4#

Putting it in trigonometric form:
#z=r(costheta+isintheta)#

#z= (sqrt2(cos((3pi)/4)+isin((3pi)/4)))^7#

When applying a power to numbers in trigonometric form, apply the power to the r, and multiply the angle with the exponent:

#z= ((sqrt2)^7(cos((3pi*7)/4)+isin((3pi*7)/4))#

#z= 8sqrt2(cos((21pi)/4)+isin((21pi)/4))#

#(21pi)/4# is coterminal with #(5pi)/4#

#z= 8sqrt2(cos((5pi)/4)+isin((5pi)/4))#

#z= 8sqrt2*cos((5pi)/4)+8sqrt2*isin((5pi)/4)#

#z= 8sqrt2*-sqrt2/2+8sqrt2*-sqrt2/2i#

#z=-8-8i#

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