Expand Using De Moivre's Theorem (3-i)^3

Problem

(3−i)3

Solution

1. Identify the complex number z=3−i and the power n=3

2. Calculate the modulus r of the complex number using the formula r=√(,a2+b2)

r=√(,3+(−1)2)

r=√(,9+1)

r=√(,10)

3. Find the argument θ using tan(θ)=b/a Since the point (3,−1) is in the fourth quadrant, we use the inverse tangent.

θ=arctan((−1)/3)

4. Apply De Moivre's Theorem, which states zn=rn*(cos(n*θ)+i*sin(n*θ))

(3−i)3=(√(,10))3*(cos(3*θ)+i*sin(3*θ))

5. Use the triple angle identities cos(3*θ)=4*cos3(θ)−3*cos(θ) and sin(3*θ)=3*sin(θ)−4*sin3(θ)

6. Determine the values of cos(θ) and sin(θ) from the original complex number.

cos(θ)=3/√(,10)

sin(θ)=(−1)/√(,10)

7. Substitute these values into the triple angle identities.

cos(3*θ)=4*(3/√(,10))3−3*(3/√(,10))

cos(3*θ)=108/(10√(,10))−9/√(,10)

cos(3*θ)=(108−90)/(10√(,10))

cos(3*θ)=18/(10√(,10))

8. Calculate the sine component.

sin(3*θ)=3*((−1)/√(,10))−4*((−1)/√(,10))3

sin(3*θ)=(−3)/√(,10)+4/(10√(,10))

sin(3*θ)=(−30+4)/(10√(,10))

sin(3*θ)=(−26)/(10√(,10))

9. Combine all parts to find the final expanded form.

(3−i)3=10√(,10)*(18/(10√(,10))+i(−26)/(10√(,10)))

(3−i)3=18−26*i

Final Answer

(3−i)3=18−26*i

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