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100+ Free Mathematics Specialised Level 4 Practice Questions
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Master TASC Mathematics Specialised Level 4 (MTS415118) with 100 exam-aligned practice questions covering complex numbers, 3D vectors, advanced integration, differential equations, proof techniques, and matrix algebra. These practice questions are an English-language multiple-choice study aid for revising course knowledge and are not an official TASC paper or a simulation of the written external examination format.
Sample Mathematics Specialised Level 4 Practice Questions
Try these sample questions to test your Mathematics Specialised Level 4 exam readiness. Each question includes a detailed explanation. Start the interactive quiz above for the full 100+ question experience with AI tutoring.
1What is the modulus of the complex number \(z = 3 - 4i\)?
A.\(\sqrt{7}\)
B.\(5\)
C.\(7\)
D.\(25\)
Explanation: The modulus of a complex number \(z = a + bi\) is calculated using \(|z| = \sqrt{a^2 + b^2}\). Substituting \(a = 3\) and \(b = -4\) gives \(|z| = \sqrt{3^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5\).
2Which of the following represents the complex number \(z = 1 + i\) in polar form \(r \text{cis}(\theta)\)?
A.\(\text{cis}\left(\frac{\pi}{4}\right)\)
B.\(2 \text{cis}\left(\frac{\pi}{4}\right)\)
C.\(\sqrt{2} \text{cis}\left(\frac{\pi}{4}\right)\)
D.\(\sqrt{2} \text{cis}\left(\frac{\pi}{2}\right)\)
Explanation: The modulus is \(r = \sqrt{1^2 + 1^2} = \sqrt{2}\). The principal argument is \(\theta = \arctan(1/1) = \frac{\pi}{4}\). Thus, \(z = \sqrt{2} \text{cis}\left(\frac{\pi}{4}\right)\).
3Express the complex number \(z = 2 \text{cis}\left(\frac{\pi}{3}\right)\) in Cartesian form \(a + bi\).
A.\(1 + i\sqrt{3}\)
B.\(\sqrt{3} + i\)
C.\(1 - i\sqrt{3}\)
D.\(2 + 2i\sqrt{3}\)
Explanation: Expanding polar form gives \(z = 2\left(\cos\frac{\pi}{3} + i\sin\frac{\pi}{3}\right) = 2\left(\frac{1}{2} + i\frac{\sqrt{3}}{2}\right) = 1 + i\sqrt{3}\).
4What is the complex conjugate of \(z = -2 + 5i\)?
A.\(2 - 5i\)
B.\(2 + 5i\)
C.\(-2 - 5i\)
D.\(-2 + 5i\)
Explanation: The complex conjugate of \(z = a + bi\) is \(\bar{z} = a - bi\). Negating only the imaginary part of \(z = -2 + 5i\) yields \(\bar{z} = -2 - 5i\).
5Simplify the powers of the imaginary unit \(i^{23}\).
A.\(1\)
B.\(-1\)
C.\(i\)
D.\(-i\)
Explanation: Since powers of \(i\) repeat with period 4 (\(i^1=i, i^2=-1, i^3=-i, i^4=1\)), we divide 23 by 4 to get remainder 3. Thus \(i^{23} = (i^4)^5 \cdot i^3 = 1^5 \cdot (-i) = -i\).
6Simplify the product of complex numbers \((2 + 3i)(1 - 2i)\).
A.\(8 - i\)
B.\(8 + i\)
C.\(-4 - i\)
D.\(2 - 6i\)
Explanation: Expanding using FOIL gives \((2+3i)(1-2i) = 2(1) - 4i + 3i - 6i^2\). Since \(i^2 = -1\), this becomes \(2 - i - 6(-1) = 2 - i + 6 = 8 - i\).
7Use De Moivre's theorem to evaluate \((1 + i)^8\).
A.\(8\)
B.\(16\)
C.\(16i\)
D.\(32\)
Explanation: First convert \(1+i\) to polar form: \(r = \sqrt{2}\), \(\theta = \frac{\pi}{4}\). By De Moivre's theorem, \((1+i)^8 = (\sqrt{2})^8 \text{cis}\left(8 \cdot \frac{\pi}{4}\right) = 16 \text{cis}(2\pi) = 16(1 + 0i) = 16\).
8Evaluate the quotient \(\frac{6 \text{cis}\left(\frac{5\pi}{6}\right)}{2 \text{cis}\left(\frac{\pi}{3}\right)}\) and express the answer in Cartesian form.
A.\(3i\)
B.\(-3i\)
C.\(3\)
D.\(\frac{3\sqrt{3}}{2} + \frac{3}{2}i\)
Explanation: Dividing complex numbers in polar form gives \(r = \frac{6}{2} = 3\) and \(\theta = \frac{5\pi}{6} - \frac{\pi}{3} = \frac{5\pi - 2\pi}{6} = \frac{3\pi}{6} = \frac{\pi}{2}\). Thus, \(3 \text{cis}\left(\frac{\pi}{2}\right) = 3(0 + i) = 3i\).
9Find all solutions to the polynomial equation \(z^3 = 8i\).
A.\(\sqrt{3} + i,\, -\sqrt{3} + i,\, -2i\)
B.\(2i,\, -2i,\, 2\)
C.\(2 \text{cis}\left(\frac{\pi}{3}\right),\, 2 \text{cis}(\pi),\, 2 \text{cis}\left(\frac{5\pi}{3}\right)\)
D.\(\sqrt{3} - i,\, -\sqrt{3} - i,\, 2i\)
Explanation: Write \(8i = 8 \text{cis}\left(\frac{\pi}{2}\right)\). By De Moivre's theorem for roots, \(z_k = 2 \text{cis}\left(\frac{\pi/2 + 2k\pi}{3}\right)\) for \(k=0,1,2\).
For \(k=0\): \(2 \text{cis}\left(\frac{\pi}{6}\right) = \sqrt{3} + i\).
For \(k=1\): \(2 \text{cis}\left(\frac{5\pi}{6}\right) = -\sqrt{3} + i\).
For \(k=2\): \(2 \text{cis}\left(\frac{9\pi}{6}\right) = 2 \text{cis}\left(\frac{3\pi}{2}\right) = -2i\).
10Describe the locus of points in the Argand plane defined by the equation \(|z - (2 + 3i)| = 4\).
A.A line passing through \((2, 3)\) with slope 4.
B.A circle centered at \((2, 3)\) with radius 4.
C.A circle centered at \((-2, -3)\) with radius 4.
D.An ellipse centered at the origin with semi-major axis 4.
Explanation: The equation \(|z - z_0| = r\) represents a circle in the Argand plane centered at \(z_0\) with radius \(r\). Here, \(z_0 = 2 + 3i\) corresponding to the point \((2, 3)\), and radius \(r = 4\).
About the Mathematics Specialised Level 4 Practice Questions
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