Convert to Trigonometric Form (3-2i)^8

Problem

(3−2*i)^8

Solution

1. Identify the complex number z=a+b*i inside the power, where a=3 and b=−2

2. Calculate the modulus r using the formula r=√(,a^2+b^2)

r=√(,3^2+(−2)^2)

r=√(,9+4)

r=√(,13)

3. Determine the argument θ using tan(θ)=b/a Since the point (3,−2) is in the fourth quadrant, θ=arctan((−2)/3)

θ=arctan(−2/3)

4. Express the base z in trigonometric form r*(cos(θ)+i*sin(θ))

z=√(,13)*(cos(arctan(−2/3))+i*sin(arctan(−2/3)))

5. Apply De Moivre's Theorem, which states that z^n=r^n*(cos(n*θ)+i*sin(n*θ))

(3−2*i)^8=(√(,13))^8*(cos(8*θ)+i*sin(8*θ))

6. Simplify the modulus (√(,13))^8=(13^(1/2))^8=13^4

13^4=28561

Final Answer

(3−2*i)^8=28561*(cos(8*arctan(−2/3))+i*sin(8*arctan(−2/3)))

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