Convert to Trigonometric Form (2+8i)-(6+9i)

Problem

(2+8*i)−(6+9*i)

Solution

1. Subtract the complex numbers by combining the real parts and the imaginary parts separately.

(2−6)+(8−9)*i

−4−i

2. Identify the real part a=−4 and the imaginary part b=−1 to find the modulus r

r=√(,a2+b2)

r=√(,(−4)2+(−1)2)

r=√(,16+1)

r=√(,17)

3. Determine the argument θ using the arctangent function. Since the point (−4,−1) is in the third quadrant, add π (or 180 to the result of arctan(b/a)

θ=arctan((−1)/(−4))+π

θ=arctan(0.25)+π

θ≈3.39* radians

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

√(,17)*(cos(3.39)+i*sin(3.39))

Final Answer

(2+8*i)−(6+9*i)=√(,17)*(cos(3.39)+i*sin(3.39))

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