A complex number is given in Cartesian form. What has to be done before de Moivre's theorem can be used on it?

Convert it into modulus-argument form or exponential form first, because the theorem is stated in terms of a modulus and an argument . The answer can then be converted back into Cartesian form if the question asks for it in that form.

A LevelFurther MathsEdexcelCore PureFlashcardsComplex NumbersExponential Form & de Moivre's Theorem

Exponential Form & de Moivre's Theorem (Edexcel A Level Further Maths: Core Pure): Flashcards

Exam code: 9FM0

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FrontExponential Form
Complete Euler's relation, which links the exponential and trigonometric functions:
$e^{i θ} =______+ i______$

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Cards in this collection (36)

Complete Euler's relation, which links the exponential and trigonometric functions:
$e^{i θ} =______+ i______$
The completed relation is:
$e^{i θ} = \cos θ + i \sin θ$
This is what raising $e$ to an imaginary power is defined to mean, and it is the bridge between exponentials and trigonometry.
How is a complex number with modulus $r$ and argument $θ$ written in exponential form?
It is written $z = r e^{i θ}$ .
This follows from the polar form $z = r (\cos θ + i \sin θ)$ by replacing the whole bracket with $e^{i θ}$ .
True or False?
For every real value of $θ$ the number $e^{i θ}$ has modulus $1$ .
True.
Euler's relation gives $e^{i θ} = \cos θ + i \sin θ$ , whose modulus is $\sqrt{\cos^{2} θ + \sin^{2} θ} = 1$ .
So as $θ$ varies, $e^{i θ}$ travels round the unit circle rather than growing the way a real exponential does.
What are $e^{i π}$ and $e^{\frac{π}{2} i}$ equal to as ordinary complex numbers?
$e^{i π} = - 1$ , since $\cos π = - 1$ and $\sin π = 0$ , which is often written as $e^{i π} + 1 = 0$ .
$e^{\frac{π}{2} i} = i$ , since $\cos \frac{π}{2} = 0$ and $\sin \frac{π}{2} = 1$ .
For $z_{1} = r_{1} e^{i θ_{1}}$ and $z_{2} = r_{2} e^{i θ_{2}}$ complete the product and the quotient:
$z_{1} z_{2} =______and \frac{z_{1}}{z_{2}} =______$
The completed results are:
$z_{1} z_{2} = r_{1} r_{2} e^{i (θ_{1} + θ_{2})} and \frac{z_{1}}{z_{2}} = \frac{r_{1}}{r_{2}} e^{i (θ_{1} - θ_{2})}$
Both are the ordinary index laws applied to $e^{i θ}$ , with the moduli handled separately.
Why does the index law $e^{i θ_{1}} \times e^{i θ_{2}} = e^{i (θ_{1} + θ_{2})}$ agree with what you already know about arguments?
Multiplying complex numbers adds their arguments, and the index law adds the powers, so the two statements say exactly the same thing.
That agreement is the main reason exponential form is convenient: the behaviour of the arguments is built into the index laws rather than being a separate rule.
If $z = r e^{i θ}$ what is $z^{*}$ in exponential form?
The conjugate is $z^{*} = r e^{- i θ}$ : you simply change the sign of the argument.
The modulus $r$ is unchanged, because $a + b i$ and $a - b i$ are the same distance from the origin.
Why does $e^{2 k π i} = 1$ hold whenever $k$ is an integer?
Because $\cos (2 k π) = 1$ and $\sin (2 k π) = 0$ for every integer $k$ , so Euler's relation gives $1 + 0 i$ .
Adding a whole number of turns to an argument returns to exactly the same point, so all of these exponentials name the number $1$ .
A complex number is given in Cartesian form. What has to be done before de Moivre's theorem can be used on it?
Convert it into modulus-argument form or exponential form first, because the theorem is stated in terms of a modulus and an argument.
The answer can then be converted back into Cartesian form if the question asks for it in that form.
True or False?
To find $z^{5}$ you multiply the modulus by $5$ and multiply the argument by $5$ .
False.
The argument is multiplied by $5$ , but the modulus is raised to the power $5$ .
So a number with modulus $2$ and argument $θ$ gives a fifth power with modulus $32$ , not $10$ .
To find $z^{3}$ for $z = 2 (\cos \frac{π}{5} - i \sin \frac{π}{5})$ what has to be done first?
Rewrite it with a plus sign as $2 (\cos (- \frac{π}{5}) + i \sin (- \frac{π}{5}))$ , so that the argument is $- \frac{π}{5}$ .
de Moivre's theorem then gives $z^{3} = 8 (\cos (- \frac{3 π}{5}) + i \sin (- \frac{3 π}{5}))$ .
For $z = r (\cos θ + i \sin θ)$ complete the reciprocal:
$\frac{1}{z} =______(\cos (______) + i \sin (______))$
The completed reciprocal is:
$\frac{1}{z} = \frac{1}{r} (\cos (- θ) + i \sin (- θ))$
It is de Moivre's theorem with an index of $- 1$ , which inverts the modulus and reverses the argument.
How can $z^{- n}$ be written using $\cos n θ$ and $\sin n θ$ rather than negative angles?
Use the identities $\cos (- n θ) = \cos n θ$ and $\sin (- n θ) = - \sin n θ$ .
These turn the result into $z^{- n} = r^{- n} (\cos n θ - i \sin n θ)$ , which is often the tidier form to quote.
Why is de Moivre's theorem better than expanding ${(2 + 3 i)}^{8}$ with the binomial theorem?
The binomial expansion gives nine terms, each needing a power of $i$ to be simplified, and the work grows quickly as the index rises.
In modulus-argument form the index is dealt with in a single step, whatever its size.
Why do the powers of a complex number sometimes repeat in a cycle?
It happens when the modulus is $1$ : every power then also has modulus $1$ , so all the powers stay on the unit circle and only the argument changes.
With the modulus fixed the arguments can come back round to where they started, and from that point onwards the powers repeat.
To express $\cos 5 θ$ as a polynomial in $\cos θ$ , what is the first move?
Write $\cos 5 θ + i \sin 5 θ$ as ${(\cos θ + i \sin θ)}^{5}$ , using de Moivre's theorem in reverse.
Expanding that bracket binomially and comparing real parts then gives $\cos 5 θ$ , while comparing imaginary parts gives $\sin 5 θ$ .
Complete the rule about which multiple-angle formulae can be written using one trig function only:
$\cos k θ$ can always be written using $\cos θ$ alone, but $\sin k θ$ can be written using $\sin θ$ alone only when $k$ is $______$ and for even $k$ a factor of $______$ is left over.
The completed rule is: $\cos k θ$ can always be written using $\cos θ$ alone, but $\sin k θ$ can be written using $\sin θ$ alone only when $k$ is odd and for even $k$ a factor of $\cos θ$ is left over.
So $\sin 5 θ$ has a pure $\sin θ$ form, while $\sin 4 θ$ does not.
Comparing real parts has left $\cos 5 θ$ in terms of both $\cos θ$ and $\sin θ$ . How do you finish the proof?
Use $\sin^{2} θ + \cos^{2} θ = 1$ to replace every power of $\sin θ$ by an expression in $\cos θ$ .
This always works because only even powers of $\sin θ$ survive in the real part, and every even power is a power of $\sin^{2} θ$ .
Once you have expansions for $\sin k θ$ and $\cos k θ$ , how do you get $\tan k θ$ in terms of $\tan θ$ alone?
Form the quotient $\tan k θ = \frac{\sin k θ}{\cos k θ}$ and divide every term by the highest power of $\cos θ$ .
That turns each term into a power of $\tan θ$ or $\sec θ$ , and $1 + \tan^{2} θ = \sec^{2} θ$ then removes the $\sec θ$ terms.
Complete the exponential expressions for cosine and sine:
$\cos k θ =______(e^{i k θ} + e^{- i k θ}) and \sin k θ =______(e^{i k θ} - e^{- i k θ})$
The completed expressions are:
$\cos k θ = \frac{1}{2} (e^{i k θ} + e^{- i k θ}) and \sin k θ = \frac{1}{2 i} (e^{i k θ} - e^{- i k θ})$
They come from adding and subtracting $e^{i k θ} = \cos k θ + i \sin k θ$ and $e^{- i k θ} = \cos k θ - i \sin k θ$ .
What is the point of rewriting $\sin^{5} θ$ in terms of $\sin θ$ , $\sin 3 θ$ and $\sin 5 θ$ ?
A power of a trig function is awkward to integrate, whereas a sum of multiple angles integrates term by term.
So $\sin^{5} θ \equiv \frac{1}{16} (10 \sin θ - 5 \sin 3 θ + \sin 5 θ)$ turns one difficult integral into three standard ones.
True or False?
Rewriting $\sin^{4} θ$ in terms of multiple angles gives an expression made of $\sin n θ$ terms.
False.
Because the index $4$ is even, the result is built from $\cos n θ$ terms instead.
A power of sine reduces to sines only when its index is odd, whereas a power of cosine always reduces to cosines.
When you expand ${(e^{i θ} + e^{- i θ})}^{k}$ why do the terms pair up so neatly?
The expansion is symmetric, so any term $A e^{i n θ}$ always has a partner $A e^{- i n θ}$ .
Each pair collapses to $2 A \cos n θ$ , and when $k$ is even one unpaired constant term is left in the middle.
In a trig series question, why is it useful to let $C$ be the cosine sum and $S$ the sine sum and then work with $C + i S$ ?
Each pair $\cos k θ + i \sin k θ$ becomes $e^{i k θ}$ , so $C + i S$ is a single geometric series in $e^{i θ}$ .
Summing it once gives both answers at the same time: $C$ is the real part of the sum and $S$ is the imaginary part.
After summing a geometric series in $e^{i θ}$ the denominator has the form $A - B e^{i n θ}$ . How do you make it real?
Multiply the numerator and the denominator by $A - B e^{- i n θ}$ .
The denominator then becomes $A^{2} + B^{2} - 2 A B \cos n θ$ , because the two middle terms combine using $e^{i n θ} + e^{- i n θ} = 2 \cos n θ$ .
When can the infinite sum $S_{\infty} = \frac{a}{1 - z}$ be used for a geometric series of complex numbers?
Only when $|z| < 1$ , exactly as for a real geometric series, with the modulus measuring the size of the common ratio.
For a finite series $S_{n} = \frac{a (1 - z^{n})}{1 - z}$ is used instead, and that needs no condition at all.
How do you find the square roots of $5 + 12 i$ without converting to polar form?
Set ${(c + d i)}^{2} = 5 + 12 i$ , expand the left-hand side, and compare real and imaginary parts to get $c^{2} - d^{2} = 5$ and $2 c d = 12$ .
Solving those two equations simultaneously gives the roots $3 + 2 i$ and $- 3 - 2 i$ .
In the square-root method, why does eliminating one unknown lead to a quartic that is easy to solve?
Rearranging $2 c d = b$ to $c = \frac{b}{2 d}$ and substituting into $c^{2} - d^{2} = a$ produces $4 d^{4} + 4 a d^{2} - b^{2} = 0$ once the fraction is cleared.
That quartic contains only $d^{4}$ and $d^{2}$ , so the substitution $u = d^{2}$ turns it into an ordinary quadratic.
For $z = r (\cos θ + i \sin θ)$ complete the formula for its $n$ th roots:
$z^{\frac{1}{n}} = r^{\frac{1}{n}} (\cos (\frac{θ +______}{n}) + i \sin (\frac{θ +______}{n}))$
The completed formula is:
$z^{\frac{1}{n}} = r^{\frac{1}{n}} (\cos (\frac{θ + 2 π k}{n}) + i \sin (\frac{θ + 2 π k}{n}))$
Taking $k = 0, 1, 2, \dots, n - 1$ in turn produces the $n$ distinct roots.
Why do the two square roots of a complex number always differ only by a sign?
If $w$ is a square root then so is $- w$ , since ${(- w)}^{2} = w^{2}$ .
A quadratic equation has exactly two roots, so $w$ and $- w$ are the only two there can be.
Why does adding $2 π k$ to the argument give $n$ different $n$ th roots rather than just one?
Adding a whole number of turns to an argument leaves the original number unchanged, so $z$ can be written with infinitely many different arguments.
Dividing each of them by $n$ gives angles that differ by $\frac{2 π}{n}$ , and only $n$ of those are distinct before the pattern comes back round.
True or False?
The equation $z^{3} = 8$ has only one solution.
False.
It has three solutions, because a cubic equation has three roots.
Only $z = 2$ is real; the other two are non-real and come from taking $k = 1$ and $k = 2$ in the $n$ th root formula.
Where do the $n$ roots of a complex number $r e^{i θ}$ lie on an Argand diagram?
They all lie on a circle centred at the origin with radius $\sqrt[n]{r}$ .
Their arguments are equally spaced $\frac{2 π}{n}$ apart, so joining consecutive roots gives a regular polygon with $n$ sides.
Complete these facts about the $n$ th roots of unity, where $ω = e^{\frac{2 π}{n} i}$ :
The roots are the powers of $ω$ from $ω^{0}$ up to $ω^{______}$ , and for $n \geq 2$ they add up to $______$ in total.
The completed facts are: the roots are the powers of $ω$ from $ω^{0}$ up to $ω^{n - 1}$ , and for $n \geq 2$ they add up to $0$ in total.
The sum is zero because the roots sit symmetrically around the origin, so as vectors they cancel out.
You have found one root $α$ of the equation $z^{n} = r e^{i θ}$ . How do you get the rest quickly?
Multiply $α$ by each $n$ th root of unity in turn, giving $α, α ω, α ω^{2}, \dots, α ω^{n - 1}$ .
Each multiplication by $ω$ leaves the modulus alone and adds $\frac{2 π}{n}$ to the argument, stepping round to the next root.
A regular hexagon is centred at the origin and one vertex is known. How can complex numbers give the other five vertices?
Treat the known vertex $(x, y)$ as the complex number $x + y i$ and multiply it by each sixth root of unity.
In exponential form that simply adds $\frac{2 π}{6}$ to the argument each time, and converting the results back to Cartesian form gives the coordinates.

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