Complex Numbers - IB Mathematics Analysis and Approaches

1. The equation $z^4 - 2z^3 + 7z^2 - 4z + 10 = 0$ has a root $z_1 = 1-2i$. Find a root of the equation that is purely imaginary.
   Answer: i\sqrt{2} The polynomial has real coefficients, so if $z_1 = 1-2i$ is a root, its conjugate $z_2 = 1+2i$ must also be a root. This means that $(z - (1-2i))(z - (1+2i))$ is a factor of the polynomial. This factor is $z^2 - (z_1+z_2)z + z_1z_2$. Sum of roots: $z_1+z_2 = (1-2i)+(1+2i) = 2$. Product of roots: $z_1z_2 = (1-2i)(1+2i) = 1^2 - (2i)^2 = 1 - (-4) = 5$. So, $z^2 - 2z + 5$ is a factor. We can find the other factor by polynomial long division: $(z^4 - 2z^3 + 7z^2 - 4z + 10) \div (z^2 - 2z + 5)$. The division gives $z^2+2$. So, $z^4 - 2z^3 + 7z^2 - 4z + 10 = (z^2 - 2z + 5)(z^2+2)$. The remaining roots are the solutions to $z^2+2=0$. $z^2 = -2 \implies z = \pm\sqrt{-2} = \pm i\sqrt{2}$. These two roots, $i\sqrt{2}$ and $-i\sqrt{2}$, are purely imaginary. Therefore, one such root is $i\sqrt{2}$.
2. A complex number $z$ satisfies the equation $|z - 3i| = 2|z|$. The locus of points representing $z$ in the complex plane is a circle. Find the centre of this circle.
   Answer: -i Let $z = x+iy$, where $x, y \in \mathbb{R}$. Substitute this into the given equation: $|(x+iy) - 3i| = 2|x+iy|$ $|x + i(y-3)| = 2|x+iy|$ Now, apply the definition of the modulus: $|a+bi| = \sqrt{a^2+b^2}$. $\sqrt{x^2 + (y-3)^2} = 2\sqrt{x^2+y^2}$ Square both sides to eliminate the square roots: $x^2 + (y-3)^2 = 4(x^2+y^2)$ $x^2 + y^2 - 6y + 9 = 4x^2 + 4y^2$ Rearrange the terms to get the equation of a circle: $3x^2 + 3y^2 + 6y - 9 = 0$ Divide the entire equation by 3: $x^2 + y^2 + 2y - 3 = 0$ To find the centre, we complete the square for the $y$ terms: $x^2 + (y^2 + 2y + 1) - 1 - 3 = 0$ $x^2 + (y+1)^2 = 4$ This is the equation of a circle with centre $(0, -1)$ and radius 2. In the complex plane, the centre $(0, -1)$ corresponds to the complex number $0 - 1i = -i$.
3. Let $\omega$ be a non-real cube root of unity. Find the value of $(1-\omega+\omega^2)(1+\omega-\omega^2)$.
   Answer: 4 The cube roots of unity are the solutions to $z^3=1$, or $z^3-1=0$. This factors as $(z-1)(z^2+z+1)=0$. The roots are $z=1$ and the roots of $z^2+z+1=0$. Since $\omega$ is a non-real root, it must satisfy $\omega^2+\omega+1=0$. From this identity, we can derive two useful relations: 1) $1+\omega^2 = -\omega$ 2) $1+\omega = -\omega^2$ Now we substitute these into the expression we want to evaluate: $(1-\omega+\omega^2)(1+\omega-\omega^2)$ For the first factor: $(1+\omega^2) - \omega = (-\omega) - \omega = -2\omega$. For the second factor: $(1+\omega) - \omega^2 = (-\omega^2) - \omega^2 = -2\omega^2$. So the product is $(-2\omega)(-2\omega^2) = 4\omega^3$. Since $\omega$ is a cube root of unity, we know that $\omega^3=1$. Therefore, the value of the expression is $4(1)=4$.
4. Find the general form of all complex numbers $z$ that satisfy the equation $z^2 + |z|^2 = 0$.
   Answer: $z = ki$ for any $k \in \mathbb{R}$ When you encounter equations involving both complex numbers and their modulus (absolute value), you need to work with both the algebraic and geometric properties of complex numbers. Let's write $z = a + bi$ where $a, b \in \mathbb{R}$. Then $|z|^2 = a^2 + b^2$, so our equation becomes: $(a + bi)^2 + a^2 + b^2 = 0$ Expanding the left side: $a^2 + 2abi + (bi)^2 + a^2 + b^2 = 0$ $a^2 + 2abi - b^2 + a^2 + b^2 = 0$ $2a^2 + 2abi = 0$ $2a(a + bi) = 0$ This gives us $a = 0$ or $a + bi = 0$. If $a + bi = 0$, then $z = 0$. If $a = 0$, then $z = bi$ for any real $b$, which we can write as $z = ki$ where $k \in \mathbb{R}$. Let's verify: if $z = ki$, then $z^2 = -k^2$ and $|z|^2 = k^2$, so $z^2 + |z|^2 = -k^2 + k^2 = 0$ ✓ Choice A is correct - it captures all purely imaginary numbers (including $z = 0$ when $k = 0$). Choice B is wrong because it only includes zero, missing all other purely imaginary solutions. Choice C is wrong because real numbers $z = k$ would give $k^2 + k^2 = 2k^2 = 0$, which only works for $k = 0$. Choice D is wrong because $z = k(1+i)$ gives $2k^2i + 2k^2 \neq 0$ for $k \neq 0$. Study tip: When solving complex equations with modulus, separate real and imaginary parts systematically - this reveals the geometric constraint that solutions must lie on specific lines or curves in the complex plane.
5. The complex numbers $z_1 = 3+2i$ and $z_2 = -1+4i$ represent two adjacent vertices of a square in the complex plane. Which of the following complex numbers could represent another vertex of the square?
   Answer: -3 Let the vertices be A($\small z_1$) and B($\small z_2$). The vector from A to B is represented by the complex number $v = z_2 - z_1 = (-1+4i) - (3+2i) = -4+2i$. The other two vertices, C and D, form the square ABCD. The vector BC must be perpendicular to AB and have the same length. We can find the vector BC by rotating the vector BA by $\pm 90^{\circ}$ ($\pm \frac{\pi}{2}$) about B. The vector BA is $-v = 4-2i$. Rotation by $90^{\circ}$ counter-clockwise corresponds to multiplication by $i$. Rotation by $90^{\circ}$ clockwise corresponds to multiplication by $-i$. \textbf{Case 1:} Rotate BA by $90^{\circ}$ counter-clockwise. Vector BC $= i \times (z_1-z_2) = i(4-2i) = 4i - 2i^2 = 2+4i$. The position of vertex C is $z_3 = z_2 + (2+4i) = (-1+4i) + (2+4i) = 1+8i$. \textbf{Case 2:} Rotate BA by $90^{\circ}$ clockwise. Vector BC $= -i \times (z_1-z_2) = -i(4-2i) = -4i + 2i^2 = -2-4i$. The position of vertex C is $z_3 = z_2 + (-2-4i) = (-1+4i) + (-2-4i) = -3$. So, two possible locations for a third vertex are $1+8i$ and $-3$. The fourth vertex would be $z_4 = z_1 + \vec{BC}$. In Case 1, $z_4 = (3+2i)+(2+4i) = 5+6i$. In Case 2, $z_4 = (3+2i)+(-2-4i) = 1-2i$. The possible vertices are $1+8i$, $5+6i$, $-3$, and $1-2i$. From the options, $-3$ is a possible vertex.
6. A complex number $z$ has modulus 4 and argument $\frac{2\pi}{3}$. Find $z - 2\bar{z}$ in the form $a+bi$.
   Answer: 2 + 6i\sqrt{3} First, we express $z$ in Cartesian form $a+bi$. Given $|z|=4$ and $\arg(z) = \frac{2\pi}{3}$. $z = |z|(\cos(\arg(z)) + i\sin(\arg(z)))$ $z = 4(\cos(\frac{2\pi}{3}) + i\sin(\frac{2\pi}{3}))$ We know that $\cos(\frac{2\pi}{3}) = -\frac{1}{2}$ and $\sin(\frac{2\pi}{3}) = \frac{\sqrt{3}}{2}$. So, $z = 4(-\frac{1}{2} + i\frac{\sqrt{3}}{2}) = -2 + 2i\sqrt{3}$. Next, we find the conjugate of $z$, $\bar{z}$. $\bar{z} = -2 - 2i\sqrt{3}$. Now, we calculate the expression $z - 2\bar{z}$: $z - 2\bar{z} = (-2 + 2i\sqrt{3}) - 2(-2 - 2i\sqrt{3})$ $= -2 + 2i\sqrt{3} + 4 + 4i\sqrt{3}$ Combine the real and imaginary parts: $= (-2+4) + (2\sqrt{3}+4\sqrt{3})i$ $= 2 + 6i\sqrt{3}$.
7. Let $z = 1-i$. Find $z^{10}$ in Cartesian form.
   Answer: -32i To find a high power of a complex number, it is best to convert it to polar or Euler form. For $z=1-i$: The modulus is $|z| = \sqrt{1^2+(-1)^2} = \sqrt{2}$. The argument is $\arg(z) = \arctan(\frac{-1}{1}) = -\frac{\pi}{4}$ (since $z$ is in the fourth quadrant). So, in Euler form, $z = \sqrt{2}e^{-i\pi/4}$. Now, we use De Moivre's theorem to find $z^{10}$: $$z^{10} = (\sqrt{2}e^{-i\pi/4})^{10} = (\sqrt{2})^{10} e^{-i(10\pi/4)} = 2^5 e^{-i(5\pi/2)}$$ We have $z^{10} = 32e^{-i(5\pi/2)}$. To convert back to Cartesian form, we find an angle coterminal with $-\frac{5\pi}{2}$ in the range $(-\pi, \pi]$. $-\frac{5\pi}{2} = -2\pi - \frac{\pi}{2}$, so the angle is equivalent to $-\frac{\pi}{2}$. $$z^{10} = 32(\cos(-\frac{\pi}{2}) + i\sin(-\frac{\pi}{2})) = 32(0 + i(-1)) = -32i$$
8. Find the sum of the infinite geometric series $1 + \frac{1+i}{2} + (\frac{1+i}{2})^2 + (\frac{1+i}{2})^3 + \dots$.
   Answer: 1+i This is an infinite geometric series with first term $a=1$ and common ratio $r = \frac{1+i}{2}$. For the series to converge, the modulus of the common ratio must be less than 1 ($|r|<1$). $|r| = |\frac{1+i}{2}| = \frac{|1+i|}{2} = \frac{\sqrt{1^2+1^2}}{2} = \frac{\sqrt{2}}{2}$. Since $\sqrt{2} \approx 1.414$, $|r| \approx 0.707 < 1$, so the series converges. The sum to infinity of a convergent geometric series is given by the formula $S = \frac{a}{1-r}$. $$S = \frac{1}{1 - \frac{1+i}{2}} = \frac{1}{\frac{2 - (1+i)}{2}} = \frac{1}{\frac{1-i}{2}} = \frac{2}{1-i}$$ To express this in the form $x+yi$, we multiply the numerator and denominator by the conjugate of the denominator: $$S = \frac{2}{1-i} \times \frac{1+i}{1+i} = \frac{2(1+i)}{1^2 - i^2} = \frac{2(1+i)}{1 - (-1)} = \frac{2(1+i)}{2} = 1+i$$
9. A complex number $z$ satisfies the equation $2z + i\bar{z} = 5-i$. Find $z$.
   Answer: \frac{11}{3} - \frac{7}{3}i Let $z = x+iy$, where $x, y \in \mathbb{R}$. Then its conjugate is $\bar{z} = x-iy$. Substitute these into the equation: $2(x+iy) + i(x-iy) = 5-i$ Distribute the terms: $2x + 2iy + ix - i^2y = 5-i$ Since $i^2 = -1$, this becomes: $2x + 2iy + ix + y = 5-i$ Group the real and imaginary parts on the left side: $(2x+y) + i(x+2y) = 5-i$ Now, equate the real and imaginary parts from both sides of the equation: Real part: $2x+y = 5$ (1) Imaginary part: $x+2y = -1$ (2) We have a system of two linear equations. From (1), we can write $y = 5-2x$. Substitute this into (2): $x + 2(5-2x) = -1$ $x + 10 - 4x = -1$ $-3x = -11 \implies x = \frac{11}{3}$ Now substitute the value of $x$ back into the expression for $y$: $y = 5 - 2(\frac{11}{3}) = 5 - \frac{22}{3} = \frac{15-22}{3} = -\frac{7}{3}$ So, $z = x+iy = \frac{11}{3} - \frac{7}{3}i$.
10. A polynomial $P(z) = z^3 + az^2 + bz + 15$ has real coefficients $a$ and $b$. Given that $z_1 = 2+i$ is a root of $P(z)$, find the value of $a$.
    Answer: -1 Since the polynomial $P(z)$ has real coefficients, if $z_1 = 2+i$ is a root, then its complex conjugate $z_2 = 2-i$ must also be a root. For a cubic polynomial, there must be a third root, $z_3$. According to Vieta's formulas, the product of the roots is $z_1z_2z_3 = -\frac{15}{1} = -15$. First, we find the product of the known roots: $z_1z_2 = (2+i)(2-i) = 2^2 - i^2 = 4 - (-1) = 5$. Now we can find the third root: $5 \cdot z_3 = -15 \implies z_3 = -3$. Also from Vieta's formulas, the sum of the roots is $z_1+z_2+z_3 = -\frac{a}{1} = -a$. Summing the three roots: $(2+i) + (2-i) + (-3) = 4 - 3 = 1$. Therefore, $-a = 1$, which means $a = -1$.
11. The locus of a complex number $z$ is given by $\arg(z-2) - \arg(z+2i) = \frac{\pi}{2}$. Which of the following describes this locus?
    Answer: A major arc of a circle The given equation can be rewritten using the property of arguments: $\arg\left(\frac{z-2}{z+2i}\right) = \frac{\pi}{2}$. Let $w = \frac{z-2}{z+2i}$. The condition $\arg(w) = \frac{\pi}{2}$ means that $w$ is a purely imaginary number with a positive imaginary part. Let $w=ki$ for some $k>0$. Let $z = x+iy$. Then $w = \frac{(x-2)+iy}{x+i(y+2)}$. We set the real part of $w$ to zero. The real part of a quotient $\frac{a+bi}{c+di}$ is $\frac{ac+bd}{c^2+d^2}$. For this to be zero, we only need $ac+bd=0$. Here, $a=x-2, b=y, c=x, d=y+2$. So, $(x-2)x + y(y+2) = 0$. $x^2-2x + y^2+2y = 0$. Completing the square gives: $(x^2-2x+1) + (y^2+2y+1) - 1 - 1 = 0$, which is $(x-1)^2+(y+1)^2=2$. This is the equation of a circle with centre $(1,-1)$ and radius $\sqrt{2}$. However, we also have the condition that $w$ has a positive imaginary part. The imaginary part of the quotient is $\frac{bc-ad}{c^2+d^2}$. We need $bc-ad>0$. $yx - (x-2)(y+2) > 0$ $xy - (xy+2x-2y-4) > 0$ $-2x+2y+4 > 0 \implies y-x+2 > 0 \implies y > x-2$. This inequality restricts the solution to a portion of the circle. The locus starts at $z=-2i$ and ends at $z=2$ (these points are excluded). The circle passes through these two points. The arc that satisfies $y>x-2$ is the major arc of the circle connecting these two points.
12. Let the complex number $z$ be given by $z = \frac{2+ai}{1-2i}$, where $a \in \mathbb{R}$. If the real part of $z$ is zero, find the value of $a$.
    Answer: 1 To find the real part of $z$, we first write $z$ in the form $x+yi$ by multiplying the numerator and denominator by the conjugate of the denominator, which is $1+2i$. $$z = \frac{2+ai}{1-2i} \times \frac{1+2i}{1+2i} = \frac{(2+ai)(1+2i)}{1^2 + (-2)^2} = \frac{2+4i+ai+2ai^2}{1+4} = \frac{(2-2a)+(4+a)i}{5}$$ So, $z = \frac{2-2a}{5} + i\frac{4+a}{5}$. The real part of $z$ is $\text{Re}(z) = \frac{2-2a}{5}$. We are given that $\text{Re}(z) = 0$, so we set the real part equal to zero and solve for $a$: $$\frac{2-2a}{5} = 0 \implies 2-2a = 0 \implies 2 = 2a \implies a=1$$
13. By considering the binomial expansion of $(\cos\theta + i\sin\theta)^4$, which of the following is an expression for $\cos(4\theta)$ in terms of $\cos\theta$?
    Answer: 8\cos^4\theta - 8\cos^2\theta + 1 This question tests your understanding of De Moivre's theorem and how to use binomial expansion to derive trigonometric identities. When you see $(\cos\theta + i\sin\theta)^n$, think about using both De Moivre's theorem and the binomial theorem to find relationships between trigonometric functions. By De Moivre's theorem, $(\cos\theta + i\sin\theta)^4 = \cos(4\theta) + i\sin(4\theta)$. Using the binomial theorem to expand the left side: $(\cos\theta + i\sin\theta)^4 = \cos^4\theta + 4i\cos^3\theta\sin\theta + 6i^2\cos^2\theta\sin^2\theta + 4i^3\cos\theta\sin^3\theta + i^4\sin^4\theta$ Since $i^2 = -1$, $i^3 = -i$, and $i^4 = 1$, this becomes: $\cos^4\theta + 4i\cos^3\theta\sin\theta - 6\cos^2\theta\sin^2\theta - 4i\cos\theta\sin^3\theta + \sin^4\theta$ Grouping real and imaginary parts: $(\cos^4\theta - 6\cos^2\theta\sin^2\theta + \sin^4\theta) + i(4\cos^3\theta\sin\theta - 4\cos\theta\sin^3\theta)$ The real part equals $\cos(4\theta)$. Using $\sin^2\theta = 1 - \cos^2\theta$: $\cos(4\theta) = \cos^4\theta - 6\cos^2\theta(1-\cos^2\theta) + (1-\cos^2\theta)^2$ Expanding and simplifying: $\cos(4\theta) = 8\cos^4\theta - 8\cos^2\theta + 1$ Answer A is correct. Answer B has the wrong coefficient for the $\cos^2\theta$ term. Answer C has incorrect coefficients for both the $\cos^4\theta$ and $\cos^2\theta$ terms. Answer D shows the unsimplified form before substituting for $\sin^2\theta$. Remember: When using De Moivre's theorem with binomial expansion, always substitute $\sin^2\theta = 1 - \cos^2\theta$ to express everything in terms of $\cos\theta$ only.
14. How many distinct solutions does the equation $z|z| - 2z + i = 0$ have?
    Answer: 2 Rearrange the equation to isolate $z$: $z(|z|-2) = -i$. From this form, we can see that $z = \frac{-i}{|z|-2}$. Since the right side is a real multiple of $-i$, $z$ must be a purely imaginary number. Let $z=iy$ for some real number $y$. The modulus is $|z| = |iy| = |y|$. Substitute $z=iy$ and $|z|=|y|$ into the original equation: $iy|y| - 2(iy) + i = 0$ Since $i \neq 0$, we can divide the entire equation by $i$: $y|y| - 2y + 1 = 0$. We must consider two cases based on the sign of $y$. **Case 1: $y \ge 0$** In this case, $|y|=y$. The equation becomes: $y(y) - 2y + 1 = 0 \implies y^2 - 2y + 1 = 0$ $(y-1)^2 = 0 \implies y=1$. Since $y=1$ satisfies the condition $y \ge 0$, this gives a valid solution $z = i(1) = i$. **Case 2: $y < 0$** In this case, $|y|=-y$. The equation becomes: $y(-y) - 2y + 1 = 0 \implies -y^2 - 2y + 1 = 0$ $y^2 + 2y - 1 = 0$. Using the quadratic formula to solve for $y$: $y = \frac{-2 \pm \sqrt{2^2 - 4(1)(-1)}}{2(1)} = \frac{-2 \pm \sqrt{8}}{2} = \frac{-2 \pm 2\sqrt{2}}{2} = -1 \pm \sqrt{2}$. We have two potential values for $y$: $y_1 = -1+\sqrt{2} \approx 0.414$ and $y_2 = -1-\sqrt{2} \approx -2.414$. Since this case requires $y<0$, we must discard $y_1$. The only valid solution in this case is $y_2 = -1-\sqrt{2}$, which gives the solution $z = i(-1-\sqrt{2})$. In total, we have two distinct solutions: $z=i$ and $z = (-1-\sqrt{2})i$.
15. Let $z_1 = 1+i\sqrt{3}$ and $z_2 = 1-i$. Find the principal argument of $\frac{z_1^4}{z_2^2}$.
    Answer: -\frac{\pi}{6} We use the property $\arg(\frac{w_1}{w_2}) = \arg(w_1) - \arg(w_2)$ and $\arg(z^n) = n\arg(z)$. First, find the arguments of $z_1$ and $z_2$. For $z_1 = 1+i\sqrt{3}$, it is in the first quadrant. $\arg(z_1) = \arctan(\frac{\sqrt{3}}{1}) = \frac{\pi}{3}$. For $z_2 = 1-i$, it is in the fourth quadrant. $\arg(z_2) = \arctan(\frac{-1}{1}) = -\frac{\pi}{4}$. Now, we find the argument of the expression: $\arg(\frac{z_1^4}{z_2^2}) = \arg(z_1^4) - \arg(z_2^2) = 4\arg(z_1) - 2\arg(z_2)$ $= 4(\frac{\pi}{3}) - 2(-\frac{\pi}{4}) = \frac{4\pi}{3} + \frac{\pi}{2} = \frac{8\pi}{6} + \frac{3\pi}{6} = \frac{11\pi}{6}$. This is a valid argument, but it is not the principal argument, which must be in the interval $(-\pi, \pi]$. To find the principal argument, we add or subtract multiples of $2\pi$. $\frac{11\pi}{6} - 2\pi = \frac{11\pi}{6} - \frac{12\pi}{6} = -\frac{\pi}{6}$. Since $-\pi < -\frac{\pi}{6} \le \pi$, the principal argument is $-\frac{\pi}{6}$.
16. Find the area of the polygon whose vertices in the complex plane are given by the roots of the equation $z^4 = -16$.
    Answer: 8 First, we find the roots of $z^4 = -16$. We express -16 in polar form. The modulus is 16 and the argument is $\pi$. So, $-16 = 16e^{i\pi}$. Let $z = re^{i\theta}$. Then $z^4 = r^4e^{i4\theta}$. Comparing with $16e^{i(\pi+2k\pi)}$, we have: $r^4 = 16 \implies r=2$. $4\theta = \pi + 2k\pi \implies \theta = \frac{\pi+2k\pi}{4}$ for $k=0, 1, 2, 3$. The arguments are: $\theta_0 = \frac{\pi}{4}$, $\theta_1 = \frac{3\pi}{4}$, $\theta_2 = \frac{5\pi}{4}$, $\theta_3 = \frac{7\pi}{4}$. The four roots all lie on a circle of radius 2 centred at the origin. They are equally spaced angularly, with a separation of $\frac{2\pi}{4} = \frac{\pi}{2}$. Therefore, the vertices form a square inscribed in a circle of radius 2. The polygon can be seen as four isosceles triangles with the origin as a common vertex. Each triangle has two sides of length 2 (the radius) and the angle between them is $\frac{\pi}{2}$. The area of one such triangle is $A_{triangle} = \frac{1}{2}ab\sin C = \frac{1}{2}(2)(2)\sin(\frac{\pi}{2}) = 2(1) = 2$. Since there are four such triangles, the total area of the square is $4 \times 2 = 8$.
17. Find the value of the sum $S = \sum_{n=1}^{101} i^n$.
    Answer: i The sum is $S = i + i^2 + i^3 + i^4 + \dots + i^{101}$. The powers of $i$ form a cycle of length 4: $i, -1, -i, 1$. The sum of each cycle is $i - 1 - i + 1 = 0$. We can group the terms in sets of four. The number of terms is 101. $101 = 4 \times 25 + 1$. This means we have 25 full cycles of four terms, plus one extra term. The sum of the first 100 terms is $\sum_{n=1}^{100} i^n = 25 \times (i+i^2+i^3+i^4) = 25 \times 0 = 0$. So, the sum $S$ is just the last term, $i^{101}$. To evaluate $i^{101}$, we use the remainder when 101 is divided by 4. $101 = 4 \times 25 + 1$, so $i^{101} = (i^4)^{25} \cdot i^1 = 1^{25} \cdot i = i$. Therefore, $S = i$.
18. The transformation $T$ of the complex plane is defined by $w = (1+i)z + 2$. Which sequence of transformations maps a point $z$ to $w$?
    Answer: A rotation by $\frac{\pi}{4}$ counter-clockwise about the origin, an enlargement of scale factor $\sqrt{2}$ from the origin, followed by a translation by $\begin{pmatrix} 2 \\ 0 \end{pmatrix}$. The transformation is $w = (1+i)z + 2$. This is of the form $w = az+b$, which represents a rotation and dilation, followed by a translation. The order of operations is multiplication first, then addition. 1. **Multiplication by $a=1+i$:** We write $1+i$ in polar form. $|1+i| = \sqrt{1^2+1^2} = \sqrt{2}$ and $\arg(1+i) = \frac{\pi}{4}$. So, $1+i = \sqrt{2}e^{i\pi/4}$. Multiplication by this complex number corresponds to an enlargement (dilation) by a scale factor of $\sqrt{2}$ from the origin and a rotation by $\frac{\pi}{4}$ counter-clockwise about the origin. These two can be performed in any order. 2. **Addition of $b=2$:** Adding 2 corresponds to a translation by the vector $\begin{pmatrix} 2 \\ 0 \end{pmatrix}$. So the full transformation is a rotation and enlargement, followed by a translation. Choice B correctly describes this sequence. Choice A has the wrong order: translating first would lead to $w = (1+i)(z+2) = (1+i)z + 2+2i$, which is a different transformation.
19. How many distinct non-zero complex numbers $z$ satisfy the equation $z^2 = \bar{z}$?
    Answer: 3 Let's solve the equation $z^2 = \bar{z}$ using polar form. Let $z = re^{i\theta}$. Then $\bar{z} = re^{-i\theta}$. The equation becomes $(re^{i\theta})^2 = re^{-i\theta}$, which simplifies to $r^2e^{i2\theta} = re^{-i\theta}$. We can equate the moduli and the arguments. Equating moduli: $r^2 = r \implies r^2 - r = 0 \implies r(r-1) = 0$. This gives $r=0$ or $r=1$. If $r=0$, then $z=0$. This is one solution. If $r=1$, we equate the arguments: $e^{i2\theta} = e^{-i\theta}$. This implies that $2\theta = -\theta + 2k\pi$ for some integer $k$. $3\theta = 2k\pi \implies \theta = \frac{2k\pi}{3}$. We find the distinct values for $\theta$ by taking $k=0, 1, 2$. For $k=0$, $\theta=0$. This gives the solution $z = 1e^{i0} = 1$. For $k=1$, $\theta=\frac{2\pi}{3}$. This gives the solution $z = e^{i2\pi/3} = -\frac{1}{2} + i\frac{\sqrt{3}}{2}$. For $k=2$, $\theta=\frac{4\pi}{3}$. This gives the solution $z = e^{i4\pi/3} = -\frac{1}{2} - i\frac{\sqrt{3}}{2}$. For $k=3$, we get $\theta = 2\pi$, which is the same as $\theta=0$. The solutions are $0, 1, e^{i2\pi/3}, e^{i4\pi/3}$. The question asks for the number of non-zero solutions. There are 3 non-zero solutions.
20. Given $z = \cos\theta + i\sin\theta$, which of the following expressions is equivalent to $\frac{z^2-1}{z^2+1}$?
    Answer: i\tan\theta We can express $z$ in Euler form as $z=e^{i\theta}$. Then $z^2 = e^{i2\theta}$. Substitute this into the expression: $$\frac{z^2-1}{z^2+1} = \frac{e^{i2\theta}-1}{e^{i2\theta}+1}$$ To simplify, we can factor out $e^{i\theta}$ from both the numerator and the denominator: $$\frac{e^{i\theta}(e^{i\theta}-e^{-i\theta})}{e^{i\theta}(e^{i\theta}+e^{-i\theta})} = \frac{e^{i\theta}-e^{-i\theta}}{e^{i\theta}+e^{-i\theta}}$$ Using the definitions $\cos\theta = \frac{e^{i\theta}+e^{-i\theta}}{2}$ and $\sin\theta = \frac{e^{i\theta}-e^{-i\theta}}{2i}$, we have: $e^{i\theta}-e^{-i\theta} = 2i\sin\theta$ $e^{i\theta}+e^{-i\theta} = 2\cos\theta$ Substituting these back into the expression gives: $$\frac{2i\sin\theta}{2\cos\theta} = i\frac{\sin\theta}{\cos\theta} = i\tan\theta$$

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