Convert to Trigonometric Form -5i(4-3i)^2

Problem

−5*i*(4−3*i)2

Solution

1. Expand the squared binomial (4−3*i)2 using the formula (a−b)2=a2−2*a*b+b2

(4−3*i)2=16−24*i+9*i2

2. Simplify the expression by substituting i2=−1

16−24*i−9=7−24*i

3. Distribute the −5*i term into the simplified binomial.

−5*i*(7−24*i)=−35*i+120*i2

4. Substitute i2=−1 again to find the standard form a+b*i

−35*i−120=−120−35*i

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

r=√(,(−120)2+(−35)2)=√(,14400+1225)=√(,15625)=125

6. Determine the argument θ using tan(θ)=b/a Since the point (−120,−35) is in the third quadrant, add 180 (or π radians) to the reference angle.

θ=arctan((−35)/(−120))+180≈16.26+180=196.26

7. Write the result in trigonometric form r*(cos(θ)+i*sin(θ))

125*(cos(196.26)+i*sin(196.26))

Final Answer

−5*i*(4−3*i)2=125*(cos(196.26)+i*sin(196.26))

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