Simplify (1- square root of 3i)^5

Problem

(1−√(,3)*i)5

Solution

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

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

r=√(,1+(−√(,3))2)

r=√(,1+3)

r=2

3. Determine the argument θ using tan(θ)=b/a Since the point (1,−√(,3)) is in the fourth quadrant, we find the angle.

tan(θ)=(−√(,3))/1

θ=−π/3

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

z=2*(cos(−π/3)+i*sin(−π/3))

5. Apply De Moivre's Theorem, which states that zn=rn*(cos(n*θ)+i*sin(n*θ)) for n=5

z5=2*(cos(5⋅−π/3)+i*sin(5⋅−π/3))

z5=32*(cos(−(5*π)/3)+i*sin(−(5*π)/3))

6. Simplify the angle by finding its coterminal equivalent in the range [0,2*π)

−(5*π)/3+2*π=π/3

7. Evaluate the trigonometric functions at π/3

cos(π/3)=1/2

sin(π/3)=√(,3)/2

8. Distribute the modulus to find the final rectangular form.

z5=32*(1/2+i√(,3)/2)

z5=16+16√(,3)*i

Final Answer

(1−√(,3)*i)5=16+16√(,3)*i

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