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=a^2−2*a*b+b^2

(4−3*i)^2=16−24*i+9*i^2

2. Simplify the expression by substituting i^2=−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*i^2

4. Substitute i^2=−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=√(,a^2+b^2)

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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