Mastery Examination: Complete Solutions Manual for Complex Variables and Applications, 9th Edition by Brown and Churchill WITH STEP BY STEP SOLUTIONS AND IN-DEPTH RATIONALES 2026 !

Mastery Examination: Complete Solutions Manual for
Complex Variables and Applications, 9th Edition by
Brown and Churchill WITH STEP BY STEP SOLUTIONS
AND IN-DEPTH RATIONALES 2026 !


SECTION 1: COMPLEX NUMBERS AND BASIC ALGEBRAIC PROPERTIES (Questions 1-25)

Question 1

Topic: Complex Number Operations

Problem: Express (3+4i)(2-5i) in the form a+bi.



CORRECT ANSWER: 26-7i



Rationale: Using FOIL multiplication: (3)(2)=6, (3)(-5i)=-15i, (4i)(2)=8i, (4i)(-5i)=-20i² = 20.
Sum real parts: 6+20=26; imaginary parts: -15i+8i=-7i. This demonstrates the fundamental
algebraic operation of complex multiplication, where i²=-1 is crucial .



Question 2

Topic: Complex Conjugates

Problem: Find the conjugate of z = 5-3i and compute z·\bar{z}.



CORRECT ANSWER: \bar{z}=5+3i, z·\bar{z}=34



Rationale: The conjugate changes the sign of the imaginary part. The product z·\bar{z}
= (5-3i)(5+3i) = 25+15i-15i-9i² = 25+9 = 34, which equals |z|². This property is fundamental
for computing moduli and rationalizing denominators .



Question 3

Topic: Modulus and Argument

,Problem: Find |z| and Arg(z) for z = -2+2i.



CORRECT ANSWER: |z|=2√2, Arg(z)=3π/4



Rationale: |z| = √(a²+b²) = √(4+4) = 2√2. The argument lies in quadrant II since a<0, b>0,
so Arg(z) = π - arctan(|b/a|) = π - arctan(1) = π - π/4 = 3π/4. Understanding modulus
and argument is essential for polar form representation .



Question 4

Topic: Polar Form

Problem: Convert z = 3-3i to polar form.



CORRECT ANSWER: z = 3√2(cos(7π/4)+i sin(7π/4))



Rationale: |z| = √(9+9)=3√2. The point lies in quadrant IV, so θ = -π/4 = 7π/4. Polar form z
= r(cos θ + i sin θ) = 3√2(cos(7π/4)+i sin(7π/4)) .



Question 5

Topic: De Moivre's Theorem

Problem: Evaluate (1+i)⁶ using De Moivre's Theorem.



CORRECT ANSWER: -8i



Rationale: First convert to polar form: 1+i = √2(cos(π/4)+i sin(π/4)). By De Moivre's
Theorem: (1+i)⁶ = (√2)⁶(cos(6π/4)+i sin(6π/4)) = 8(cos(3π/2)+i sin(3π/2)) = 8(0-i) = -8i .



Question 6

Topic: Roots of Complex Numbers

Problem: Find all cube roots of -8.

, CORRECT ANSWER: 2e^(iπ/3), 2, 2e^(i5π/3)



Rationale: -8 = 8e^(iπ). The cube roots are given by z_k = ∛8 · e^(i(π+2πk)/3) for k=0,1,2.
Thus: k=0: 2e^(iπ/3), k=1: 2e^(iπ)=2, k=2: 2e^(i5π/3). These three points form an
equilateral triangle on the circle |z|=2 .



Question 7

Topic: Quadratic Equations with Complex Coefficients

Problem: Solve z² + (2+3i)z - (1+5i) = 0.



CORRECT ANSWER: z = 2-2i and z = -4-i



Rationale: Using the quadratic formula: z = [-(2+3i) ± √((2+3i)² + 4(1+5i))]/2. The
discriminant simplifies to 4+12i-9+4+20i = -1+32i. The square root of -1+32i is
±(√(√1025+1)/2 + i√(√1025-1)/2). Simplifying yields the two roots .



Question 8

Topic: Triangle Inequality

Problem: Show that |z₁+z₂| ≤ |z₁|+|z₂| for z₁=3+4i, z₂=12-5i.



CORRECT ANSWER: |3+4i+12-5i| = |15-i| = √226 ≈ 15.033 ≤ 5+13 = 18



Rationale: The triangle inequality states |z₁+z₂| ≤ |z₁|+|z₂|. Here |z₁| = √(9+16)=5,
|z₂| = √(144+25)=13, and |z₁+z₂| = |15-i| = √226 ≈ 15.033, which is indeed less than 18.
This inequality is fundamental in complex analysis .



Question 9

Topic: Regions in Complex Plane

Problem: Describe the region |z-2| < 3 geometrically.



CORRECT ANSWER: Open disk centered at (2,0) with radius 3

, Rationale: The inequality |z-2| < 3 represents all points whose distance from the point 2
(on the real axis) is less than 3. This forms an open circular disk centered at 2 with radius
3, not including the boundary circle .



Question 10

Topic: Argument Properties

Problem: Find Arg(z₁z₂) if Arg(z₁)=π/3 and Arg(z₂)=5π/6.



CORRECT ANSWER: 7π/6



Rationale: The argument of a product equals the sum of the arguments: Arg(z₁z₂) =
Arg(z₁) + Arg(z₂) = π/3 + 5π/6 = 2π/6 + 5π/6 = 7π/6. This property follows from the polar
form representation .



Question 11

Topic: Vector Representation

Problem: Interpret z₁-z₂ geometrically.



CORRECT ANSWER: Vector from z₂ to z₁



Rationale: In the complex plane, z₁-z₂ represents the vector with initial point z₂ and
terminal point z₁. Its magnitude |z₁-z₂| is the distance between the two points. This
geometric interpretation is fundamental for understanding limits and continuity .



Question 12

Topic: Complex Exponential Form

Problem: Express -1 - i in exponential form.



CORRECT ANSWER: √2 e^(i5π/4)