Simplify (- square root of 3+i)^6

Problem

(−√(,3)+i)^6

Solution

1. Identify the complex number in rectangular form z=a+b*i where a=−√(,3) and b=1

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

r=√(,(−√(,3))^2+1^2)

r=√(,3+1)

r=2

3. Determine the argument θ by finding the angle in the second quadrant where cos(θ)=(−√(,3))/2 and sin(θ)=1/2

θ=(5*π)/6

4. Express the complex number in polar form z=r*(cos(θ)+i*sin(θ))

z=2*(cos((5*π)/6)+i*sin((5*π)/6))

5. Apply De Moivre's Theorem, which states that z^n=r^n*(cos(n*θ)+i*sin(n*θ)) for n=6

z^6=2^6*(cos(6⋅(5*π)/6)+i*sin(6⋅(5*π)/6))

6. Simplify the exponent and the angle.

z^6=64*(cos(5*π)+i*sin(5*π))

7. Evaluate the trigonometric functions at 5*π Since 5*π is coterminal with π cos(5*π)=−1 and sin(5*π)=0

z^6=64*(−1+i(0))

z^6=−64

Final Answer

(−√(,3)+i)^6=−64

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