Question 15
De Moivre’s theorem can be expressed as
Prove De Moivre’s theorem using mathematical induction.
[6 marks]Every power comes from the one before it by the same move
Watch the orange arrow. Each time it is multiplied by one more $z$, it turns through $\theta$ and stretches by $r$. That single move is the inductive step.
Checking a few powers shows the pattern. The proof has to show that the move from $z^k$ to $z^{k+1}$ works for every $k$.
Walk through the inductive step one stage at a time
Choose $k$, $r$ and $\theta$, then step through the four stages. The black arrow is $z^k$, which the assumption tells you about. The orange arrow is $z^{k+1}$. Every line of working is the same complex number, so the value at the bottom never changes as you step through.
An index law writes the power $k+1$ as the power $k$ times one more factor. That one extra factor is what lets the assumption in.
The assumption replaces $\big(r\,\text{cis}\,\theta\big)^k$ with $r^k\,\text{cis}(k\theta)$. In a written proof, this is the line that earns the mark for using the assumption.
Expand, use $i^2=-1$, and group the real part and the imaginary part. The blue dashed lines show those two parts of $z^{k+1}$ on the axes.
By the angle sum identities, the real bracket collapses to $\cos(k\theta+\theta)$ and the imaginary bracket collapses to $\sin(k\theta+\theta)$. That is the right-hand side with $n=k+1$, and the blue spiral shows the argument $(k+1)\theta$.
With $r\lt1$ the powers spiral inwards. The proof still works, because $r^k\times r=r^{k+1}$ for any $r$.
Work through the solution one mark at a time
Try each step yourself before you reveal it. This follows the QCAA’s first method, which works from the left-hand side in the inductive step.
Show the rule is true for $n=1$, the first positive integer. Work out each side separately.
So the rule is true for $n=1$. Equivalent setting out is accepted, such as $\text{LHS}=\text{RHS}=r\,\text{cis}(\theta)$. Writing “LHS” and “RHS” is recommended but not required.
Marker: correctly proves the initial statement
Assume the rule is true for $n=k$. Write it out in full with $k$ in place of $n$.
The cis form, $\big(r\,\text{cis}(\theta)\big)^k=r^k\,\text{cis}(k\theta)$, is also accepted. Markers can still award the later marks if a bracket ends up misaligned, but careful brackets make your working much easier to follow.
Marker: correctly states a suitable assumption
Write down what you need to show for $n=k+1$.
Start from the left-hand side. Split off one factor with an index law, then use the assumption on the power $k$.
The second line is where the assumption is used. The cis form $r^k\,\text{cis}(k\theta)\,r\,\text{cis}(\theta)$ is also accepted, and follow-through marks apply if an earlier line had a slip.
Marker: uses the assumption statement
Multiply $r^k$ by $r$ and expand the two brackets. Then use $i^2=-1$ and group the real and imaginary parts.
Group them like this on purpose. With $A=k\theta$ and $B=\theta$, the blue real part has the shape of $\cos A\cos B-\sin A\sin B$ and the blue imaginary part has the shape of $\sin A\cos B+\cos A\sin B$. This mark can be implied by your later working. Equivalent expressions are accepted, such as writing each product the other way round, as in $\cos(\theta)\cos(k\theta)-\sin(\theta)\sin(k\theta)$.
Marker: expresses result in Cartesian form
Use the angle sum identities from the formula book, with $A=k\theta$ and $B=\theta$.
Each blue group from step 4 collapses into one term.
The cis form $r^{k+1}\,\text{cis}\big((k+1)\theta\big)$ is also accepted.
Marker: uses angle sum and difference identities
Finish the proof in words. Say that the rule is true for $n=k+1$, then give mathematical induction as the reason it is true for every positive integer.
Equivalent wording is accepted.
The marking guide also gives two other routes, and each earns the same six marks. Method 2 starts from the right-hand side. It expands $\cos(k\theta+\theta)$ and $\sin(k\theta+\theta)$ first, regroups using the common factor $\cos(k\theta)+i\sin(k\theta)$, and uses the assumption last. Method 3 works on both sides until they meet at the same product. In those two methods, the regrouping line takes the place of the Cartesian form mark.
Marker: completes proof and states a suitable conclusion
Each phrase of the question gave you something
The question gives you the statement and one instruction, and nothing in between. You have to plan the inductive step yourself: split off a factor so the assumption can be used, then see that expanding into Cartesian form lines the terms up with the angle sum identities. That joins polar form and Cartesian form in one argument, with no part a) to start you off. It stays complex familiar because proving De Moivre’s theorem by induction is a proof you can practise in advance, and the four-part structure is the one you have practised on sums and divisibility.
If I had to guess where you might lose marks, it would be the jump from the product straight to $\cos\big((k+1)\theta\big)$. The marking guide gives one mark for the Cartesian form and another for using the angle sum identities. The Cartesian form mark can be implied by later working, but a jump straight to the answer shows neither step. The subject report does not single out this question, but its practices to strengthen include giving students opportunities “to practise the algebraic manipulation of complex numbers in both forms”, and steps 4 and 5 are that manipulation. Also write the conclusion. A proof that stops at “= RHS” misses the last mark.
Only mark this done when you could do it without help. Reading the solution does not count.
Question wording and marking-guide steps are from the 2025 QCAA Specialist Mathematics external assessment, © State of Queensland (QCAA) 2025, licensed under CC BY 4.0, and have been adapted. Tangent Tuition is not affiliated with the QCAA.
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