Convert to Trigonometric Form (1-i)^10

Problem

(1−i)10

Solution

1. Identify the complex number z=1−i in the form a+b*i where a=1 and b=−1

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

r=√(,1+(−1)2)

r=√(,2)

3. Determine the argument θ using tan(θ)=b/a Since the point (1,−1) is in the fourth quadrant, θ=−π/4 or (7*π)/4

θ=arctan((−1)/1)

θ=−π/4

4. Write the complex number z in trigonometric form r*(cos(θ)+i*sin(θ))

z=√(,2)*(cos(−π/4)+i*sin(−π/4))

5. Apply De Moivre's Theorem, which states zn=rn*(cos(n*θ)+i*sin(n*θ)) for n=10

(1−i)10=(√(,2))10*(cos(10⋅−π/4)+i*sin(10⋅−π/4))

6. Simplify the modulus and the argument. Note that (√(,2))10=2=32 and the argument is −(10*π)/4=−(5*π)/2

(1−i)10=32*(cos(−(5*π)/2)+i*sin(−(5*π)/2))

7. Find the coterminal angle for −(5*π)/2 by adding 4*π (or 2⋅2*π to bring it into the standard range [0,2*π)

−(5*π)/2+4*π=(3*π)/2

Final Answer

(1−i)10=32*(cos((3*π)/2)+i*sin((3*π)/2))

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