16 exam-style questions. Tap Show answer for full working. Difficulty builds from Easy to Hard.
Given that \( z = 3 + 4i \), find:
(a) \( |z| \)
(b) \( \arg(z) \) in radians, to 3 significant figures
(c) \( z^* \) (the complex conjugate)
(a) \( |z| = \sqrt{3^2 + 4^2} = \sqrt{9+16} = \sqrt{25} = 5 \)
(b) \( \arg(z) = \tan^{-1}(4/3) \approx 0.927 \) radians (first quadrant)
(c) \( z^* = 3 - 4i \)
Express \( \dfrac{2+i}{1-i} \) in the form \( a + bi \), where \( a \) and \( b \) are real.
Multiply numerator and denominator by the conjugate of the denominator:
\( \dfrac{2+i}{1-i} \cdot \dfrac{1+i}{1+i} = \dfrac{(2+i)(1+i)}{1+1} = \dfrac{2 + 2i + i + i^2}{2} = \dfrac{2 + 3i - 1}{2} = \dfrac{1 + 3i}{2} \)
So \( a = \dfrac{1}{2} \), \( b = \dfrac{3}{2} \)
Solve the equation \( z^2 + 4z + 13 = 0 \), giving your answers in the form \( a + bi \).
Using the quadratic formula:
\( z = \dfrac{-4 \pm \sqrt{16 - 52}}{2} = \dfrac{-4 \pm \sqrt{-36}}{2} = \dfrac{-4 \pm 6i}{2} \)
\( z = -2 + 3i \) or \( z = -2 - 3i \)
Find the modulus and argument of \( w = -2\sqrt{3} + 2i \).
Give the argument in the interval \( (-\pi, \pi] \).
\( |w| = \sqrt{(-2\sqrt{3})^2 + 2^2} = \sqrt{12 + 4} = \sqrt{16} = 4 \)
Point is in the second quadrant: \( \arg(w) = \pi - \tan^{-1}\left(\dfrac{2}{2\sqrt{3}}\right) = \pi - \tan^{-1}\left(\dfrac{1}{\sqrt{3}}\right) = \pi - \dfrac{\pi}{6} = \dfrac{5\pi}{6} \)
The complex number \( z \) satisfies \( |z - 2i| = 3 \).
(a) Sketch the locus of \( z \) on an Argand diagram.
(b) Find the greatest and least possible values of \( |z| \).
(a) Circle centre \( (0, 2) \), radius 3.
(b) Distance from origin to centre is 2.
Greatest \( |z| = 2 + 3 = 5 \)
Least \( |z| = 3 - 2 = 1 \)
Given \( z = x + yi \) where \( x, y \in \mathbb{R} \), and \( z + 2z^* = 6 + 4i \), find the values of \( x \) and \( y \).
\( z^* = x - yi \)
\( z + 2z^* = (x + yi) + 2(x - yi) = x + yi + 2x - 2yi = 3x - yi \)
So \( 3x - yi = 6 + 4i \)
Equating real and imaginary parts: \( 3x = 6 \Rightarrow x = 2 \)
\( -y = 4 \Rightarrow y = -4 \)
Find the square roots of \( 5 + 12i \) in the form \( a + bi \), where \( a \) and \( b \) are real.
Let \( \sqrt{5+12i} = a + bi \). Then \( (a+bi)^2 = 5 + 12i \)
\( a^2 - b^2 + 2abi = 5 + 12i \)
So \( a^2 - b^2 = 5 \) and \( 2ab = 12 \Rightarrow ab = 6 \Rightarrow b = 6/a \)
\( a^2 - 36/a^2 = 5 \). Multiply by \( a^2 \): \( a^4 - 5a^2 - 36 = 0 \)
\( (a^2 - 9)(a^2 + 4) = 0 \Rightarrow a^2 = 9 \) (since \( a^2 > 0 \))
\( a = 3 \) or \( a = -3 \). If \( a = 3 \), \( b = 2 \); if \( a = -3 \), \( b = -2 \)
Square roots: \( 3 + 2i \) and \( -3 - 2i \)
The point \( P \) representing \( z \) on an Argand diagram satisfies \( \arg(z - 1) = \dfrac{\pi}{4} \).
(a) Sketch the locus of \( P \).
(b) The point also satisfies \( |z - 1| = 2\sqrt{2} \). Find the exact coordinates of the possible positions of \( P \).
(a) Half-line from \( (1, 0) \) at angle \( \pi/4 \) to the positive real axis (excluding the point itself).
(b) Intersection of the ray with the circle centre \( (1,0) \) radius \( 2\sqrt{2} \).
Point is at distance \( 2\sqrt{2} \) along direction \( \pi/4 \):
\( x = 1 + 2\sqrt{2} \cdot \cos(\pi/4) = 1 + 2\sqrt{2} \cdot \frac{\sqrt{2}}{2} = 1 + 2 = 3 \)
\( y = 0 + 2\sqrt{2} \cdot \sin(\pi/4) = 2 \)
So \( P \) is at \( (3, 2) \). (Only one point on the half-line.)
Given that \( z = 2e^{i\pi/3} \), express \( z^4 \) in the form \( a + bi \), where \( a \) and \( b \) are exact real numbers.
\( z^4 = 2^4 e^{i 4\pi/3} = 16\left(\cos\dfrac{4\pi}{3} + i\sin\dfrac{4\pi}{3}\right) \)
\( = 16\left(-\dfrac{1}{2} + i\left(-\dfrac{\sqrt{3}}{2}\right)\right) = -8 - 8\sqrt{3}\, i \)
The complex numbers \( z \) and \( w \) satisfy:
\( z + w = 4 \) and \( zw = 5 + 2i \)
Find \( z \) and \( w \) in the form \( a + bi \).
\( z \) and \( w \) are roots of \( t^2 - (z+w)t + zw = 0 \)
\( t^2 - 4t + (5 + 2i) = 0 \)
\( t = \dfrac{4 \pm \sqrt{16 - 4(5+2i)}}{2} = \dfrac{4 \pm \sqrt{16 - 20 - 8i}}{2} = \dfrac{4 \pm \sqrt{-4 - 8i}}{2} \)
Find \( \sqrt{-4-8i} \). Let \( \sqrt{-4-8i} = p + qi \):
\( p^2 - q^2 = -4 \), \( 2pq = -8 \Rightarrow pq = -4 \)
\( p^2 - 16/p^2 = -4 \). Multiply: \( p^4 + 4p^2 - 16 = 0 \)
\( (p^2 + 2)^2 - 20 = 0 \) wait — better: \( u = p^2 \), \( u^2 + 4u - 16 = 0 \)
\( u = -2 \pm \sqrt{20} = -2 \pm 2\sqrt{5} \). Take positive: \( p^2 = -2 + 2\sqrt{5} \)
\( p = \pm\sqrt{2(\sqrt{5}-1)} \), \( q = -4/p \)
Alternatively: \( \sqrt{-4-8i} = \sqrt{20}\left(\cos\theta + i\sin\theta\right)/2 \) style — exact forms:
One pair: \( z, w = 2 \pm \sqrt{1 - (5+2i)/1} \) simplified gives
\( z, w = 2 \pm \sqrt{-1 - 2i} \). Computing \( \sqrt{-1-2i} = 1 - i \) check: \( (1-i)^2 = 1 - 2i + i^2 = -2i \) — close but need \( -1-2i \).
\( (1-i)^2 = -2i \). Try \( a = 1, b = -1 \): not quite. \( (p+qi)^2 = -4-8i \).
Known result: \( \sqrt{-4-8i} = 2 - 2i \) check: \( (2-2i)^2 = 4 - 8i + 4i^2 = 4 - 8i - 4 = -8i \) — wrong.
\( \sqrt{20} \approx 4.47 \). From quadratic \( u^2 + 4u - 16 = 0 \), \( u = -2 + 2\sqrt{5} \approx 2.472 \), \( p \approx \pm 1.572 \), \( q = -4/p \approx \mp 2.544 \).
Exact: \( z, w = 2 \pm \sqrt{-1-2i} \) where \( \sqrt{-1-2i} = \sqrt{\dfrac{\sqrt{5}-1}{2}} - i\sqrt{\dfrac{\sqrt{5}+1}{2}} \) (or opposite signs).
Accept numerical / exact radical form matching the system.
On an Argand diagram, the point representing \( z \) satisfies both:
\( |z - 3| = |z + i| \) and \( \arg\left(\dfrac{z-3}{z+i}\right) = \dfrac{\pi}{2} \)
Find the exact value of \( z \).
\( |z-3| = |z+i| \) is the perpendicular bisector of \( (3,0) \) and \( (0,-1) \).
Midpoint: \( \left(\dfrac{3}{2}, -\dfrac{1}{2}\right) \). Direction of segment: \( (-3, -1) \), so perpendicular has direction \( (1, -3) \) or equation:
Let \( z = x+yi \): \( (x-3)^2 + y^2 = x^2 + (y+1)^2 \)
\( x^2 - 6x + 9 + y^2 = x^2 + y^2 + 2y + 1 \Rightarrow -6x + 9 = 2y + 1 \Rightarrow 2y = -6x + 8 \Rightarrow y = -3x + 4 \)
\( \arg\left(\dfrac{z-3}{z+i}\right) = \pi/2 \) means the vector from \( -i \) to \( z \) to the vector from 3 to \( z \) differs by \( \pi/2 \) — i.e. the directed angle is a right angle, and \( z \) lies on the circle with diameter from \( (3,0) \) to \( (0,-1) \) (Thales), specifically the arc where the oriented angle is \( +\pi/2 \).
Circle diameter \( (3,0) \) to \( (0,-1) \): centre \( \left(\dfrac{3}{2}, -\dfrac{1}{2}\right) \), radius \( \dfrac{\sqrt{10}}{2} \).
Equation: \( \left(x-\dfrac{3}{2}\right)^2 + \left(y+\dfrac{1}{2}\right)^2 = \dfrac{10}{4} \)
Intersect with \( y = -3x + 4 \): substitute and solve — the line is the perpendicular bisector, which passes through the centre, so intersection with the circle is at the two ends of the diameter… wait.
Actually \( |z-3|=|z+i| \) is the perp bisector; the arg condition selects points on the circle with that diameter where the turn is anticlockwise \( \pi/2 \). The intersection of the line (perp bisector) with that circle: the centre is on the line, so intersections are the two points at radius distance along the line from centre.
Direction of line \( y = -3x+4 \) has direction vector \( (1, -3) \), unit vector \( \dfrac{(1,-3)}{\sqrt{10}} \).
Points: \( \left(\dfrac{3}{2}, -\dfrac{1}{2}\right) \pm \dfrac{\sqrt{10}}{2} \cdot \dfrac{(1,-3)}{\sqrt{10}} = \left(\dfrac{3}{2}, -\dfrac{1}{2}\right) \pm \left(\dfrac{1}{2}, -\dfrac{3}{2}\right) \)
\( = (2, -2) \) or \( (1, 1) \)
Check arg: for \( z = 1+i \), \( \dfrac{z-3}{z+i} = \dfrac{-2+i}{1+2i} \). Multiply by conjugate: etc. One of these will give \( +\pi/2 \).
Testing \( z = 1+i \): \( z-3 = -2+i \), \( z+i = 1+2i \). Argument difference… \( \arg(-2+i) - \arg(1+2i) \). \( \arg(1+2i) = \tan^{-1}2 \), \( \arg(-2+i) = \pi - \tan^{-1}(1/2) \). Difference is \( \pi/2 \). Yes.
So \( z = 1 + i \).
Find all complex numbers \( z \) such that \( z^3 = 8i \), giving your answers in the form \( re^{i\theta} \) where \( -\pi < \theta \leq \pi \).
\( 8i = 8e^{i\pi/2} = 8e^{i(\pi/2 + 2k\pi)} \), \( k \in \mathbb{Z} \)
\( z = 2 e^{i(\pi/2 + 2k\pi)/3} = 2 e^{i(\pi/6 + 2k\pi/3)} \)
For \( k = 0 \): \( 2e^{i\pi/6} \)
For \( k = 1 \): \( 2e^{i(\pi/6 + 2\pi/3)} = 2e^{i(5\pi/6)} \)
For \( k = 2 \): \( 2e^{i(\pi/6 + 4\pi/3)} = 2e^{i(3\pi/2)} \) but \( 3\pi/2 > \pi \), so use \( 2e^{-i\pi/2} \) (same ray)
Answers: \( 2e^{i\pi/6} \), \( 2e^{i 5\pi/6} \), \( 2e^{-i\pi/2} \)
Given \( z = 1 - i\sqrt{3} \), express \( z \) in the form \( r(\cos\theta + i\sin\theta) \) where \( r > 0 \) and \( -\pi < \theta \leq \pi \).
\( |z| = \sqrt{1 + 3} = 2 \)
Fourth quadrant: \( \theta = -\tan^{-1}(\sqrt{3}/1) = -\pi/3 \)
So \( z = 2\left(\cos(-\pi/3) + i\sin(-\pi/3)\right) \)
Let \( z = x + yi \) where \( x, y \in \mathbb{R} \). Find the locus of \( z \) such that \( |z - 1| = 2|z + i| \).
Give the Cartesian equation of the locus.
\( |z-1|^2 = 4|z+i|^2 \)
\( (x-1)^2 + y^2 = 4\left(x^2 + (y+1)^2\right) \)
\( x^2 - 2x + 1 + y^2 = 4x^2 + 4(y^2 + 2y + 1) \)
\( x^2 - 2x + 1 + y^2 = 4x^2 + 4y^2 + 8y + 4 \)
\( 0 = 3x^2 + 3y^2 + 2x + 8y + 3 \)
Divide by 3: \( x^2 + y^2 + \dfrac{2}{3}x + \dfrac{8}{3}y + 1 = 0 \)
Complete the square: \( \left(x + \dfrac{1}{3}\right)^2 + \left(y + \dfrac{4}{3}\right)^2 = \dfrac{1}{9} + \dfrac{16}{9} - 1 = \dfrac{8}{9} \)
Circle centre \( \left(-\dfrac{1}{3}, -\dfrac{4}{3}\right) \), radius \( \dfrac{2\sqrt{2}}{3} \).
Solve the equation \( z^2 - (3+i)z + (2+2i) = 0 \), giving your answers in the form \( a + bi \).
Discriminant: \( (3+i)^2 - 4(2+2i) = 9 + 6i + i^2 - 8 - 8i = 9 + 6i - 1 - 8 - 8i = 0 - 2i \)
\( \sqrt{-2i} \): write \( -2i = 2e^{-i\pi/2} \), so square roots \( \sqrt{2}\,e^{-i\pi/4} \) and \( \sqrt{2}\,e^{i 3\pi/4} \)
\( \sqrt{2}\left(\cos(-\pi/4)+i\sin(-\pi/4)\right) = 1 - i \)
\( \sqrt{2}\left(\cos(3\pi/4)+i\sin(3\pi/4)\right) = -1 + i \)
Thus \( z = \dfrac{3+i \pm (1-i)}{2} \)
\( z = \dfrac{4}{2} = 2 \) or \( z = \dfrac{2 + 2i}{2} = 1 + i \)
(Using the other root gives the same pair.)
The complex number \( z \) satisfies \( \arg\left(\dfrac{z-2}{z+2}\right) = \dfrac{\pi}{3} \).
(a) Describe the locus of \( z \) geometrically.
(b) Find the Cartesian equation of the locus.
(c) Find the least possible value of \( |z| \).
(a) Arc of a circle (part of the circle through the points \( 2 \) and \( -2 \) on the real axis) on which the directed angle from the chord joining \( -2 \) to \( 2 \) is \( \pi/3 \). (The major or minor arc is determined by the sign of the argument.)
(b) Let \( z = x+yi \). Then \( \arg(z-2) - \arg(z+2) = \pi/3 \).
Using the formula for \( \arg\left(\dfrac{z-a}{z-b}\right) \): the locus is an arc of the circle
\( (x-2)(x+2) + y^2 = \sqrt{3}\, y \cdot 4 \) (from geometry or expansion), more systematically:
\( \tan(\arg(z-2)-\arg(z+2)) = \tan(\pi/3) = \sqrt{3} \), leading after clearing to
\( x^2 - 4 + y^2 = 2\sqrt{3}\, y \) wait — correct expansion gives circle centre on the imaginary axis:
Centre \( (0, 2/\sqrt{3}) \), radius \( 4/\sqrt{3} \) (or equivalent form \( x^2 + (y - 2/\sqrt{3})^2 = 16/3 \)).
(The part of the circle above the real axis for positive argument \( \pi/3 \).)
(c) Least \( |z| \) is the distance from origin to the nearest point on the arc, which is \( |c| - r \) if origin outside, but here centre is at distance \( 2/\sqrt{3} \) and radius \( 4/\sqrt{3} \), so origin is inside; the least modulus on the arc (above real axis, excluding the points \( \pm 2 \)) approaches the smaller intersection with the imaginary axis or the vertical tangent — minimum occurs at the point on the circle closest to origin that lies on the valid arc: distance from O to centre is \( 2/\sqrt{3} \), so min distance is \( 4/\sqrt{3} - 2/\sqrt{3} = 2/\sqrt{3} \).