Complex Numbers — Diagnostic Tests

Unit Tests

UT-1: Modulus, Argument, and Argand Diagrams

Question: (a) Plot $z_1 = 3 + 4i$ and $z_2 = -2 + i$ on an Argand diagram. (b) Calculate $|z_1|$, $\arg(z_1)$, $|z_2|$, and $\arg(z_2)$. (c) Express $z_1$ and $z_2$ in modulus-argument form $r(\cos\theta + i\sin\theta)$. (d) Calculate $z_1 z_2$ and $z_1/z_2$ using modulus-argument form.

Solution:

(b) $|z_1| = \sqrt{9 + 16} = 5$. $\arg(z_1) = \arctan(4/3) \approx 0.927$ rad $= 53.1^\circ$. $|z_2| = \sqrt{4 + 1} = \sqrt{5}$. $\arg(z_2) = \pi - \arctan(1/2) \approx \pi - 0.464 = 2.678$ rad $= 153.4^\circ$.

(c) $z_1 = 5(\cos 0.927 + i\sin 0.927)$. $z_2 = \sqrt{5}(\cos 2.678 + i\sin 2.678)$.

(d) $z_1 z_2$: $|z_1 z_2| = 5\sqrt{5}$. $\arg(z_1 z_2) = 0.927 + 2.678 = 3.605$ rad. $z_1 z_2 = 5\sqrt{5}(\cos 3.605 + i\sin 3.605)$.

$z_1/z_2$: $|z_1/z_2| = 5/\sqrt{5} = \sqrt{5}$. $\arg(z_1/z_2) = 0.927 - 2.678 = -1.751$ rad. $z_1/z_2 = \sqrt{5}(\cos(-1.751) + i\sin(-1.751))$.

UT-2: De Moivre's Theorem

Question: (a) State De Moivre's theorem. (b) Use De Moivre's theorem to find $(1 + i)^8$. (c) Find the four fourth roots of $-16$. (d) Verify that the product of all four roots equals $-16$.

Solution:

(a) De Moivre's theorem: $[r(\cos\theta + i\sin\theta)]^n = r^n(\cos n\theta + i\sin n\theta)$.

(b) $1 + i = \sqrt{2}(\cos\pi/4 + i\sin\pi/4)$. $(1+i)^8 = (\sqrt{2})^8(\cos 2\pi + i\sin 2\pi) = 16(1 + 0i) = 16$.

(c) $-16 = 16(\cos\pi + i\sin\pi)$. Fourth roots: $r = 2$, $\theta = \frac◆LB◆\pi + 2k\pi◆RB◆◆LB◆4◆RB◆$ for $k = 0,1,2,3$.

$k=0$: $2(\cos\pi/4 + i\sin\pi/4) = 2\left(\frac◆LB◆1◆RB◆◆LB◆\sqrt{2}◆RB◆ + \frac◆LB◆i◆RB◆◆LB◆\sqrt{2}◆RB◆\right) = \sqrt{2} + \sqrt{2}i$. $k=1$: $2(\cos 3\pi/4 + i\sin 3\pi/4) = 2\left(-\frac◆LB◆1◆RB◆◆LB◆\sqrt{2}◆RB◆ + \frac◆LB◆i◆RB◆◆LB◆\sqrt{2}◆RB◆\right) = -\sqrt{2} + \sqrt{2}i$. $k=2$: $2(\cos 5\pi/4 + i\sin 5\pi/4) = 2\left(-\frac◆LB◆1◆RB◆◆LB◆\sqrt{2}◆RB◆ - \frac◆LB◆i◆RB◆◆LB◆\sqrt{2}◆RB◆\right) = -\sqrt{2} - \sqrt{2}i$. $k=3$: $2(\cos 7\pi/4 + i\sin 7\pi/4) = 2\left(\frac◆LB◆1◆RB◆◆LB◆\sqrt{2}◆RB◆ - \frac◆LB◆i◆RB◆◆LB◆\sqrt{2}◆RB◆\right) = \sqrt{2} - \sqrt{2}i$.

(d) Product $= (\sqrt{2} + \sqrt{2}i)(-\sqrt{2} + \sqrt{2}i)(-\sqrt{2} - \sqrt{2}i)(\sqrt{2} - \sqrt{2}i)$.

First pair: $(\sqrt{2} + \sqrt{2}i)(-\sqrt{2} + \sqrt{2}i) = -2 + 2i - 2i + 2i^2 = -2 - 2 = -4$. Second pair: $(-\sqrt{2} - \sqrt{2}i)(\sqrt{2} - \sqrt{2}i) = -2 + 2i - 2i + 2i^2 = -2 - 2 = -4$. Product $= (-4)(-4) = 16 \ne -16$.

Let me recalculate. Actually: the product of the first pair is $(\sqrt{2})^2 + (\sqrt{2})^2 = 2 + 2 = 4$... No. $(a+bi)(a-bi) = a^2 + b^2$. But these are not conjugate pairs in the right order.

Let me just multiply: $(-16)^{1/4}$ has 4 roots whose product should be the constant term with sign: for $z^4 + 16 = 0$, the product of roots $= 16$ (by Vieta's, with sign for even degree). This is correct: the product of the 4th roots of $-16$ is $(-1)^4 \times 16 = 16$... Actually, the product of the $n$th roots of $w$ is $w \cdot (-1)^{n-1}$. For $n=4$: $(-16)(-1)^3 = 16$. So the product is 16, not $-16$. The question has an error.

UT-3: Roots of Unity

Question: (a) Find all five fifth roots of unity and show them on an Argand diagram. (b) Show that the sum of all five roots is zero. (c) Express each root in the form $\cos(2\pi k/5) + i\sin(2\pi k/5)$ for $k = 0, 1, 2, 3, 4$. (d) Find the value of $1 + \omega + \omega^2 + \omega^3 + \omega^4$ where $\omega = \cos(2\pi/5) + i\sin(2\pi/5)$.

Solution:

(a) The fifth roots of unity are solutions to $z^5 = 1 = \cos 0 + i\sin 0$. They lie on the unit circle at angles $2\pi k/5$ for $k = 0, 1, 2, 3, 4$.

(b) The sum of all $n$th roots of unity is zero for $n \gt 1$. This follows from the fact that they are the roots of $z^n - 1 = 0$, and the coefficient of $z^{n-1}$ is zero (by Vieta's formulas).

(c) $z_k = \cos(2\pi k/5) + i\sin(2\pi k/5)$ for $k = 0, 1, 2, 3, 4$.

$z_0 = 1$, $z_1 = \cos(2\pi/5) + i\sin(2\pi/5)$, $z_2 = \cos(4\pi/5) + i\sin(4\pi/5)$, $z_3 = \cos(6\pi/5) + i\sin(6\pi/5)$, $z_4 = \cos(8\pi/5) + i\sin(8\pi/5)$.

(d) Since $\omega$ is a primitive 5th root of unity: $1 + \omega + \omega^2 + \omega^3 + \omega^4 = 0$ (the sum of all 5th roots of unity is zero).

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Integration Tests

IT-1: Complex Numbers and Loci (with Matrices)

Question: (a) Find the locus of points $z$ such that $|z - 3 - i| = 2|z + 1 + 2i|$. Describe the locus geometrically. (b) The transformation $w = \frac{z - i}{z + i}$ maps the upper half of the Argand diagram ($\text{Im}(z) \gt 0$) to a circle in the $w$-plane. Find the centre and radius of this circle. (c) If $z = x + iy$, express $w = \frac{z - i}{z + i}$ in terms of $x$ and $y$.

Solution:

(a) Let $z = x + iy$. $|z - 3 - i|^2 = 4|z + 1 + 2i|^2$. $(x-3)^2 + (y-1)^2 = 4[(x+1)^2 + (y+2)^2]$. $x^2 - 6x + 9 + y^2 - 2y + 1 = 4x^2 + 8x + 4 + 4y^2 + 16y + 16$. $x^2 - 6x + y^2 - 2y + 10 = 4x^2 + 8x + 4y^2 + 16y + 20$. $0 = 3x^2 + 14x + 3y^2 + 18y + 10$. $x^2 + \frac{14}{3}x + y^2 + 6y = -\frac{10}{3}$. $\left(x + \frac{7}{3}\right)^2 - \frac{49}{9} + (y+3)^2 - 9 = -\frac{10}{3}$. $\left(x + \frac{7}{3}\right)^2 + (y+3)^2 = \frac{49}{9} + 9 - \frac{10}{3} = \frac{49 + 81 - 30}{9} = \frac{100}{9}$.

The locus is a circle with centre $\left(-\frac{7}{3}, -3\right)$ and radius $\frac{10}{3}$.

(c) $w = \frac{x + iy - i}{x + iy + i} = \frac{x + i(y-1)}{x + i(y+1)} = \frac{[x + i(y-1)][x - i(y+1)]}{x^2 + (y+1)^2}$.

Numerator: $x^2 - ix(y+1) + ix(y-1) + (y-1)(y+1) = x^2 + (y^2 - 1) + i[-x(y+1) + x(y-1)] = x^2 + y^2 - 1 - 2ix$.

$w = \frac{x^2 + y^2 - 1 - 2ix}{x^2 + (y+1)^2}$.

IT-2: De Moivre and Trigonometry (with Further Algebra)

Question: (a) Use De Moivre's theorem to find expressions for $\cos 3\theta$ and $\sin 3\theta$ in terms of $\cos\theta$ and $\sin\theta$. (b) Hence solve $\cos 3\theta = 0$ for $0 \le \theta \le \pi$. (c) Find the exact value of $\cos 3\theta$ when $\cos\theta = 1/3$. (d) Show that $\cos 3\theta = 4\cos^3\theta - 3\cos\theta$.

Solution:

(a) $(\cos\theta + i\sin\theta)^3 = \cos 3\theta + i\sin 3\theta$.

Expanding using the binomial theorem: $= \cos^3\theta + 3\cos^2\theta(i\sin\theta) + 3\cos\theta(i\sin\theta)^2 + (i\sin\theta)^3$ $= \cos^3\theta + 3i\cos^2\theta\sin\theta - 3\cos\theta\sin^2\theta - i\sin^3\theta$ $= (\cos^3\theta - 3\cos\theta\sin^2\theta) + i(3\cos^2\theta\sin\theta - \sin^3\theta)$.

$\cos 3\theta = \cos^3\theta - 3\cos\theta\sin^2\theta$. $\sin 3\theta = 3\cos^2\theta\sin\theta - \sin^3\theta$.

(b) $\cos 3\theta = 0$ means $3\theta = \pi/2, 3\pi/2, 5\pi/2, ...$ So $\theta = \pi/6, \pi/2, 5\pi/6$ in $[0, \pi]$.

(c) $\cos 3\theta = 4\cos^3\theta - 3\cos\theta = 4(1/27) - 3(1/3) = 4/27 - 9/9 = 4/27 - 1 = -23/27$.

(d) Using $\sin^2\theta = 1 - \cos^2\theta$: $\cos 3\theta = \cos^3\theta - 3\cos\theta(1 - \cos^2\theta) = \cos^3\theta - 3\cos\theta + 3\cos^3\theta = 4\cos^3\theta - 3\cos\theta$.

IT-3: Complex Transformations (with Polar Coordinates)

Question: (a) The transformation $w = z^2$ maps the line $\text{Re}(z) = 1$ in the $z$-plane. Find the image in the $w$-plane. (b) If $z = 2e^{i\pi/6}$, find $w = z^4$ in modulus-argument form. (c) Find the image of the region $|z| \le 2$, $0 \le \arg(z) \le \pi/4$ under the transformation $w = z^3$. (d) A complex number $z$ satisfies $|z - 1 - 2i| = |z - 3 - 4i|$. Show that this represents a straight line and find its equation.

Solution:

(a) Let $z = 1 + iy$. $w = z^2 = (1+iy)^2 = 1 - y^2 + 2iy$. Let $w = u + iv$: $u = 1 - y^2$ and $v = 2y$. From $v = 2y$: $y = v/2$. $u = 1 - v^2/4$. This is a parabola $u = 1 - v^2/4$, opening to the left.

(b) $w = z^4 = (2e^{i\pi/6})^4 = 2^4 e^{i4\pi/6} = 16e^{i2\pi/3} = 16(\cos 2\pi/3 + i\sin 2\pi/3) = 16(-1/2 + i\sqrt{3}/2) = -8 + 8\sqrt{3}i$.

(c) Under $w = z^3$: $|w| = |z|^3 \le 8$, and $\arg(w) = 3\arg(z) \in [0, 3\pi/4]$. The image is the sector $|w| \le 8$, $0 \le \arg(w) \le 3\pi/4$.

(d) $|z - 1 - 2i|^2 = |z - 3 - 4i|^2$. Let $z = x + iy$: $(x-1)^2 + (y-2)^2 = (x-3)^2 + (y-4)^2$. $x^2 - 2x + 1 + y^2 - 4y + 4 = x^2 - 6x + 9 + y^2 - 8y + 16$. $-2x - 4y + 5 = -6x - 8y + 25$. $4x + 4y = 20$. $x + y = 5$.

This is a straight line: the perpendicular bisector of the segment joining $(1, 2)$ and $(3, 4)$.