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Modulus and argument, polar form, De Moivre's theorem, and the roots of unity as evenly spaced points on the unit circle.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
In this lesson complex numbers stop being pairs of symbols and become points, and the algebra you already know turns into geometry. The modulus is a distance and the argument is a direction, and in that language multiplication is simply stretch-and-turn: moduli multiply and arguments add. Powering follows immediately as De Moivre's theorem, and running it backwards shows why $z^n = 1$ has exactly $n$ solutions, sitting at the corners of a regular polygon on the unit circle.
You can add and multiply complex numbers in the form $a + bi$, using $i^2 = -1$, and you have just converted between Cartesian and polar coordinates. Those two skills are about to become one: a complex number is a point of the plane, and its polar coordinates turn out to be the form in which multiplication makes sense.
Complex plane: the plane with the real part across and the imaginary part up.
Modulus $|z|$: the distance from the origin, $\sqrt{a^2+b^2}$.
Argument $\arg z$: the angle anticlockwise from the positive real axis.
Polar form: $z = r(\cos\theta + i\sin\theta)$.
De Moivre's theorem: $\left[r(\cos\theta + i\sin\theta)\right]^n = r^n(\cos n\theta + i\sin n\theta)$.
Roots of unity: the $n$ solutions of $z^n = 1$.
Write $z = a + bi$ at the point $(a, b)$ and everything becomes geometry. Its modulus $|z| = \sqrt{a^2+b^2}$ is its distance from the origin, so $|3 + 4i| = 5$, and its argument is the angle anticlockwise from the positive real axis — sketch the point rather than trusting $\arctan\frac{b}{a}$, which cannot tell opposite quadrants apart. In polar form, $z = r(\cos\theta + i\sin\theta)$, so $1 + i$ is $\sqrt{2}(\cos 45^\circ + i\sin 45^\circ)$. Polar form exists because it is the form in which multiplication is simple: multiplying two complex numbers multiplies their moduli and adds their arguments. Multiplying is stretching and turning; multiplying by $i$, of modulus $1$ and argument $90^\circ$, is a pure quarter turn. Repeating that gives De Moivre's theorem, $z^n$ has modulus $r^n$ and argument $n\theta$, so $(1+i)^4$ has modulus $(\sqrt2)^4 = 4$ and argument $180^\circ$: the number $-4$. Run it backwards and you get roots: $z^n = 1$ needs modulus $1$ and argument a multiple of $\frac{360^\circ}{n}$, which is $n$ points evenly spaced round the unit circle — the $n$th roots of unity.
Another way: picture
The unit circle with the five fifth roots of unity marked, one at $1$ on the real axis and the others every $72^\circ$ round, joined into a regular pentagon: the solutions of $z^5 = 1$ are the corners of a regular polygon.
Another way: story
In Cartesian form, adding is easy and multiplying is a slog. In polar form it is the other way round: multiplying is 'stretch by this much, turn by that much', which two numbers can do one after the other without any algebra at all.
"Multiplying adds the moduli." It multiplies them, and adds the arguments. Getting these two the wrong way round is the single commonest error in polar arithmetic, and $|zw| = |z||w|$ is the fact to hold on to.
"$z^n = 1$ has one solution, $z = 1$." It has exactly $n$, evenly spaced on the unit circle. Over the complex numbers a polynomial of degree $n$ always has $n$ roots, counted properly.
Trusting the inverse tangent for the argument. $\arctan\frac{y}{x}$ cannot tell the first quadrant from the third. For $-1 - i$ it says $45^\circ$, but the argument is $225^\circ$. Sketch the point.
$1 + i$ has modulus $\sqrt{2}$ and argument $45^\circ$.
Convert to polar first.
The fourth power has modulus $(\sqrt2)^4 = 4$ and argument $4 \times 45^\circ = 180^\circ$.
Power the modulus, multiply the argument.
Modulus $4$ at $180^\circ$ is the point $(-4, 0)$, so $(1+i)^4 = -4$.
Convert back.
Modulus $1$ and argument a multiple of $\frac{360^\circ}{3} = 120^\circ$.
Roots share the circle equally.
So $0^\circ$, $120^\circ$ and $240^\circ$: the points $1$, $-\tfrac12 + \tfrac{\sqrt3}{2}i$ and $-\tfrac12 - \tfrac{\sqrt3}{2}i$, the corners of an equilateral triangle.
$-4$ has modulus $4$ and argument $180^\circ$.
A square root halves the argument and takes the square root of the modulus: modulus $2$ at $90^\circ$ and at $270^\circ$, which are $2i$ and $-2i$.
What is the modulus of $11 + 60i$?
The modulus is answer.
What is the argument of $1 + i$, in degrees, measured anticlockwise from the positive real axis and taken between $0$ and $360$?
The argument is answer degrees.
A complex number is multiplied by a number with modulus $3$ and argument $46^\circ$. What happens to its point on the complex plane?
One complex number has modulus $6$ and argument $78^\circ$; another has modulus $8$ and argument $77^\circ$. Give the modulus and argument of their product.
modulus m, argument a degrees
Use polar form to compute $(1 + i)^{2}$. Give the real part and then the imaginary part.
real part a, imaginary part b
A drawing program stores the corner $(1, 5)$ as the complex number $1 + 5i$ and rotates a shape a quarter turn anticlockwise about the origin by multiplying every corner by $i$. Where does this corner land?
(x, y)
A designer wants $4$ identical lamps spaced evenly round a circular ceiling rose, and places them at the $4$ solutions of $z^{4} = 1$. How many degrees apart are neighbouring lamps?
Neighbouring lamps are answer degrees apart.
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
How many complex solutions does $z^{10} = 1$ have?
There are answer roots.
You can work in the complex plane in polar form. Without looking: what happens to the moduli and to the arguments when two complex numbers are multiplied, and how many solutions does $z^5 = 1$ have?
8. Your turn: the square roots of $-4$, step 2