Unit 8 · Algebra · 4 teaching hours · 18 min read

Complex Numbers

What the syllabus asks for 4

Imaginary unit; algebra of complex numbers
Geometric representation
Modulus and conjugate with properties
Square root of a complex number

The equation x^{2}+1 = 0 has no real solution, because no real number squares to a negative. Rather than stop there, mathematicians invented a new number whose square is -1 and found that the resulting system — the complex numbers — obeys all the usual rules of algebra, gives every quadratic two roots, and has a clean picture as points of a plane. This unit sets up that arithmetic and that picture.

What the exam asks

Powers of i; add, subtract, multiply and divide complex numbers; plot z on an Argand diagram and find |z| and \arg z; use the conjugate properties; and find \sqrt{a+ib} by the "let \sqrt{a+ib} = x+iy" method. Cube roots of unity appear in Group B.

8.1 Imaginary unit; algebra of complex numbers

The imaginary unit

i = \sqrt{-1}, \qquad \text{so} \qquad i^{2} = -1

A complex number is z = a + ib where a, b are real. a = \mathrm{Re}(z) is the real part and b = \mathrm{Im}(z) is the imaginary part (note: b itself is real). If b = 0, z is real; if a = 0 and b \ne 0, z is purely imaginary. The set of all complex numbers is \mathbb{C}.

Two complex numbers are equal exactly when their real parts are equal and their imaginary parts are equal:

a + ib = c + id \Leftrightarrow a = c \ \text{ and } \ b = d

This single fact turns one complex equation into two real equations, and it is the engine behind almost every problem in this chapter.

Powers of i

i^{1} = i, \quad i^{2} = -1, \quad i^{3} = i^{2}\cdot i = -i, \quad i^{4} = \left(i^{2}\right)^{2} = 1

After that the pattern repeats with period 4, so i^{n} depends only on the remainder when n is divided by 4.

**Figure 1.** Multiplying by i rotates a number a quarter-turn anticlockwise, so the powers of i cycle with period 4.

Worked example 8.1

Problem. Evaluate (a) i^{2026}, (b) i^{-35}.

Solution. (a) Divide the index by 4: 2026 = 4 \times 506 + 2, so

i^{2026} = \left(i^{4}\right)^{506}\cdot i^{2} = 1^{506}\times(-1) = -1

(b) \;-35 = 4\times(-9) + 1, so i^{-35} = \left(i^{4}\right)^{-9}\cdot i^{1} = i.

Alternatively i^{-35} = \dfrac{1}{i^{35}}; since 35 = 4\times 8 + 3, i^{35} = i^{3} = -i, and \dfrac{1}{-i} = \dfrac{-1}{i}\cdot\dfrac{i}{i} = \dfrac{-i}{-1} = i. Same answer.

The four operations

Add and subtract componentwise; multiply as ordinary binomials and replace i^{2} by -1:

(a+ib)+(c+id) = (a+c) + i(b+d)

(a+ib)(c+id) = (ac-bd) + i(ad+bc)

Division is done by multiplying above and below by the conjugate of the denominator, which makes the denominator real:

\frac{a+ib}{c+id} = \frac{a+ib}{c+id}\times\frac{c-id}{c-id} = \frac{(ac+bd) + i(bc-ad)}{c^{2}+d^{2}}

\sqrt{a}\sqrt{b} = \sqrt{ab} fails for negatives

\sqrt{-4}\times\sqrt{-9} = (2i)(3i) = 6i^{2} = -6, not \sqrt{36} = 6. Always convert each negative square root to i form first, then multiply.

Worked example 8.2

Problem. Simplify (3+4i)(2-5i).

Solution. Expand as usual:

(3+4i)(2-5i) = 6 - 15i + 8i - 20i^{2}

Since i^{2} = -1, the last term is -20(-1) = +20:

= (6+20) + (-15+8)i = 26 - 7i

Worked example 8.3

Problem. Express \dfrac{2+3i}{3-4i} in the form a+ib.

Solution. Multiply numerator and denominator by 3+4i, the conjugate of the denominator:

\frac{2+3i}{3-4i}\times\frac{3+4i}{3+4i} = \frac{6+8i+9i+12i^{2}}{3^{2}+4^{2}} = \frac{6+17i-12}{25}

= \frac{-6+17i}{25} = -\frac{6}{25} + \frac{17}{25}\,i

The denominator became (3)^{2}+(4)^{2} = 25 — a real number — because (c+id)(c-id) = c^{2}+d^{2}.

Worked example 8.4

Problem. Find real x and y such that (x+iy)(3-2i) = 13+13i.

Solution. Expand the left side:

(x+iy)(3-2i) = 3x - 2xi + 3yi - 2yi^{2} = (3x+2y) + i(3y-2x)

Equate real and imaginary parts with 13+13i:

3x + 2y = 13, \qquad -2x + 3y = 13

Multiply the first by 3 and the second by 2, then add: 9x+6y = 39 and -4x+6y = 26; subtracting gives 13x = 13, so x = 1. Then 3(1)+2y = 13 gives y = 5.

x = 1, \quad y = 5

Check. (1+5i)(3-2i) = 3-2i+15i-10i^{2} = 3+13i+10 = 13+13i. Correct.

8.2 Geometric representation

Because z = a+ib is fixed by the pair of real numbers (a, b), every complex number is a point of a plane. That plane is the Argand diagram: the horizontal axis carries the real part (the real axis), the vertical axis the imaginary part (the imaginary axis).

Argand diagram, modulus, argument

z = a+ib is plotted as the point P(a, b), or as the vector \vec{OP}.

Modulus |z| = OP = \sqrt{a^{2}+b^{2}} — the distance from the origin.
Argument \arg z = \theta, the angle \vec{OP} makes with the positive real axis, measured anticlockwise. The value in (-\pi, \pi] is the principal argument.

**Figure 2.** The Argand diagram. z = 3+4i is the point (3,4); |z| = 5 is the length OP and \\theta = \\arg z is the marked angle. The conjugate \\bar{z} = 3-4i is the mirror image in the real axis.

Polar (modulus–argument) form

From the right triangle in the figure, a = r\cos\theta and b = r\sin\theta where r = |z|. Hence

Polar form

z = r\left(\cos\theta + i\sin\theta\right), \qquad r = \sqrt{a^{2}+b^{2}}, \qquad \tan\theta = \frac{b}{a}

\tan\theta = b/a alone does not fix \theta — two angles a half-turn apart share a tangent. Always check which quadrant (a,b) lies in and adjust.

Quadrant of (a,b)	Principal argument \theta, where \tan\alpha = \|b/a\|, 0 < \alpha < \pi/2
I (a>0, b>0)	\theta = \alpha
II (a<0, b>0)	\theta = \pi - \alpha
III (a<0, b<0)	\theta = -(\pi - \alpha)
IV (a>0, b<0)	\theta = -\alpha

Worked example 8.5

Problem. Express z = -1 + i\sqrt{3} in polar form.

Solution. Here a = -1, b = \sqrt{3}.

r = |z| = \sqrt{(-1)^{2}+\left(\sqrt{3}\right)^{2}} = \sqrt{1+3} = \sqrt{4} = 2

\alpha = \tan^{-1}\left(\frac{\sqrt{3}}{1}\right) = \tan^{-1}\sqrt{3} = \frac{\pi}{3}

Since a < 0 and b > 0, the point lies in the second quadrant, so

\theta = \pi - \frac{\pi}{3} = \frac{2\pi}{3} = 120^{\circ}

z = 2\left(\cos\frac{2\pi}{3} + i\sin\frac{2\pi}{3}\right)

Check. 2\cos 120^{\circ} = 2(-\tfrac12) = -1 and 2\sin 120^{\circ} = 2\left(\tfrac{\sqrt{3}}{2}\right) = \sqrt{3}. Correct.

Addition is vector addition

z_1 + z_2 is the diagonal of the parallelogram on \vec{Oz_1} and \vec{Oz_2} — exactly the triangle law for vectors. One side of a triangle is never longer than the other two together, which gives the triangle inequality

|z_1 + z_2| \le |z_1| + |z_2|

**Figure 3.** Addition follows the parallelogram law: (4+i)+(1+3i) = 5+4i. The closing side is never longer than the two others added, which is the triangle inequality.

8.3 Modulus and conjugate with properties

Conjugate

The conjugate of z = a+ib is \bar{z} = a - ib: the same point reflected in the real axis. Reflecting twice returns you to the start, so \overline{\left(\bar{z}\right)} = z.

Properties you may quote

For all z, z_1, z_2 \in \mathbb{C}:

Conjugate	Modulus
\overline{z_1 \pm z_2} = \bar{z_1} \pm \bar{z_2}	\|z_1 z_2\| = \|z_1\|\|z_2\|
\overline{z_1 z_2} = \bar{z_1}\,\bar{z_2}	\|z_1/z_2\| = \dfrac{\|z_1\|}{\|z_2\|}
\overline{\left(\dfrac{z_1}{z_2}\right)} = \dfrac{\bar{z_1}}{\bar{z_2}}	\|z\| = \|\bar{z}\| = \|-z\|
z + \bar{z} = 2\,\mathrm{Re}(z)	z\bar{z} = \|z\|^{2}
z - \bar{z} = 2i\,\mathrm{Im}(z)	\|z_1+z_2\| \le \|z_1\|+\|z_2\|

z is real \Leftrightarrow z = \bar{z}; z is purely imaginary \Leftrightarrow z = -\bar{z}.

z\bar{z} = |z|^{2}, and the reason division works

With z = a+ib,

z\bar{z} = (a+ib)(a-ib) = a^{2} - (ib)^{2} = a^{2} - i^{2}b^{2} = a^{2}+b^{2}

and |z| = \sqrt{a^{2}+b^{2}}, so z\bar{z} = |z|^{2} — always a non-negative real number. This is precisely why multiplying by the conjugate clears a complex denominator, and it gives the reciprocal formula

\frac{1}{z} = \frac{\bar{z}}{z\bar{z}} = \frac{\bar{z}}{|z|^{2}} \qquad (z \ne 0)

Worked example 8.6

Problem. Verify that |z_1 z_2| = |z_1||z_2| for z_1 = 3+4i and z_2 = 1-2i.

Solution.

|z_1| = \sqrt{9+16} = 5, \qquad |z_2| = \sqrt{1+4} = \sqrt{5}

z_1 z_2 = (3+4i)(1-2i) = 3 - 6i + 4i - 8i^{2} = 3 - 2i + 8 = 11 - 2i

|z_1 z_2| = \sqrt{121+4} = \sqrt{125} = 5\sqrt{5}

And |z_1||z_2| = 5\sqrt{5}. The two agree.

Worked example 8.7

Problem. If z = \dfrac{1+i}{1-i}, find z, |z| and \arg z.

Solution. Multiply above and below by 1+i:

z = \frac{1+i}{1-i}\times\frac{1+i}{1+i} = \frac{(1+i)^{2}}{1^{2}+1^{2}} = \frac{1+2i+i^{2}}{2} = \frac{2i}{2} = i

So z = i = 0 + 1i, giving |z| = \sqrt{0+1} = 1 and, since the point (0,1) lies on the positive imaginary axis, \arg z = \dfrac{\pi}{2} = 90^{\circ}.

Worked example 8.8

Problem. Evaluate (1+i)^{8}.

Solution. Square first and use the result repeatedly.

(1+i)^{2} = 1 + 2i + i^{2} = 2i

(1+i)^{8} = \left[(1+i)^{2}\right]^{4} = (2i)^{4} = 2^{4}i^{4} = 16 \times 1 = 16

Modulus check. |1+i| = \sqrt{2}, so |(1+i)^{8}| = \left(\sqrt{2}\right)^{8} = 16, matching the real answer 16.

Cube roots of unity

Solving z^{3} = 1, i.e. z^{3}-1 = (z-1)\left(z^{2}+z+1\right) = 0, gives z = 1 and the roots of z^{2}+z+1 = 0:

z = \frac{-1 \pm \sqrt{1-4}}{2} = \frac{-1 \pm i\sqrt{3}}{2}

Write \omega = \dfrac{-1+i\sqrt{3}}{2}. Then the three cube roots of unity are 1, \omega, \omega^{2}.

Cube roots of unity

1 + \omega + \omega^{2} = 0, \qquad \omega^{3} = 1

Each has modulus 1, and their arguments are 0^{\circ}, 120^{\circ} and 240^{\circ} — they sit at the corners of an equilateral triangle on the unit circle. The first identity is just the coefficient relation for z^{2}+z+1 = 0 together with the root z = 1.

**Figure 4.** The cube roots of unity 1, \omega, \omega^{2} lie on the unit circle 120^{\circ} apart, forming an equilateral triangle. Being symmetric about the origin's directions, they add to zero.

Worked example 8.9

Problem. If \omega is a complex cube root of unity, evaluate \left(1+\omega-\omega^{2}\right)^{2}.

Solution. From 1+\omega+\omega^{2} = 0 we get 1+\omega = -\omega^{2}. Substituting,

1+\omega-\omega^{2} = -\omega^{2}-\omega^{2} = -2\omega^{2}

\left(1+\omega-\omega^{2}\right)^{2} = \left(-2\omega^{2}\right)^{2} = 4\omega^{4} = 4\omega^{3}\cdot\omega = 4(1)\omega = 4\omega

With \omega = \dfrac{-1+i\sqrt{3}}{2} this is -2 + 2\sqrt{3}\,i.

8.4 Square root of a complex number

Every complex number has exactly two square roots, and they are negatives of each other. To find them, use the equality of complex numbers.

The method for \sqrt{a+ib}

Let \sqrt{a+ib} = x+iy with x, y real. Squaring,

a + ib = (x+iy)^{2} = x^{2} - y^{2} + 2ixy

Equating real and imaginary parts gives two real equations:

x^{2} - y^{2} = a \qquad\text{and}\qquad 2xy = b

Now use the identity \left(x^{2}+y^{2}\right)^{2} = \left(x^{2}-y^{2}\right)^{2} + (2xy)^{2}:

x^{2}+y^{2} = \sqrt{a^{2}+b^{2}} \quad (\text{positive, since } x^{2}+y^{2} > 0)

Adding and subtracting this with x^{2}-y^{2} = a gives x^{2} and y^{2}, hence x and y. The sign of 2xy = b decides which sign pairing is allowed: if b > 0, x and y have the same sign; if b < 0, opposite signs.

A faster route in the exam

For small numbers, guess-and-fit is quicker: try (x+iy)^{2} = a+ib with small integer x, y satisfying 2xy = b. For 3+4i, 2xy = 4 means xy = 2, so try (2,1): 2^{2}-1^{2} = 3. Done — the root is \pm(2+i). Always state both signs.

Worked example 8.10

Problem. Find the square roots of 3+4i.

Solution. Let \sqrt{3+4i} = x+iy. Then

x^{2}-y^{2} = 3, \qquad 2xy = 4

x^{2}+y^{2} = \sqrt{3^{2}+4^{2}} = \sqrt{25} = 5

Adding to x^{2}-y^{2} = 3: \;2x^{2} = 8, so x^{2} = 4 and x = \pm 2. Subtracting: \;2y^{2} = 2, so y^{2} = 1 and y = \pm 1.

Since 2xy = 4 > 0, x and y take the same sign. Hence

\sqrt{3+4i} = \pm(2+i)

Check. (2+i)^{2} = 4 + 4i + i^{2} = 3+4i. Correct.

Worked example 8.11

Problem. Find the square roots of -15-8i.

Solution. Let \sqrt{-15-8i} = x+iy. Then

x^{2}-y^{2} = -15, \qquad 2xy = -8

x^{2}+y^{2} = \sqrt{(-15)^{2}+(-8)^{2}} = \sqrt{225+64} = \sqrt{289} = 17

Adding: 2x^{2} = 2 \Rightarrow x = \pm 1. Subtracting: 2y^{2} = 32 \Rightarrow y = \pm 4.

Since 2xy = -8 < 0, x and y take opposite signs:

\sqrt{-15-8i} = \pm(1-4i)

Check. (1-4i)^{2} = 1 - 8i + 16i^{2} = 1 - 8i - 16 = -15-8i. Correct.

Worked example 8.12

Problem. Solve x^{2} - 4x + 13 = 0 over \mathbb{C} and comment on the roots.

Solution. D = 16 - 52 = -36 < 0, so the roots are complex.

x = \frac{4 \pm \sqrt{-36}}{2} = \frac{4 \pm 6i}{2} = 2 \pm 3i

The two roots 2+3i and 2-3i are conjugates — as they must be, because the coefficients are real. Their sum is 4 = -b/a and their product is (2+3i)(2-3i) = 4+9 = 13 = c/a, so the root–coefficient relations of Unit 7 hold just as well in \mathbb{C}.

Chapter summary

i = \sqrt{-1}, i^{2} = -1; powers of i repeat with period 4.
z = a+ib; a+ib = c+id exactly when a = c and b = d.
Divide by multiplying top and bottom by the conjugate of the denominator.
Argand diagram: z = a+ib is the point (a,b); |z| = \sqrt{a^{2}+b^{2}} and \arg z = \theta with \tan\theta = b/a, fixed by the quadrant.
Polar form z = r(\cos\theta + i\sin\theta).
\bar{z} = a-ib; z\bar{z} = |z|^{2}, z+\bar{z} = 2\,\mathrm{Re}(z), z-\bar{z} = 2i\,\mathrm{Im}(z), \dfrac{1}{z} = \dfrac{\bar{z}}{|z|^{2}}.
|z_1z_2| = |z_1||z_2|, |z_1/z_2| = \dfrac{|z_1|}{|z_2|}, |z_1+z_2| \le |z_1|+|z_2|.
Cube roots of unity: 1, \omega, \omega^{2} with \omega^{3} = 1 and 1+\omega+\omega^{2} = 0.
\sqrt{a+ib} = x+iy: solve x^{2}-y^{2} = a, 2xy = b, x^{2}+y^{2} = \sqrt{a^{2}+b^{2}}; the sign of b fixes the sign pairing.

Practice questions

Group A — Multiple choice (1 mark each)

The value of i^{10} is [1]a1b-1cid-i
The modulus of 3+4i is [1]a7b5c\sqrt{7}d25
The conjugate of -2+5i is [1]a2+5ib-2-5ic2-5id5-2i
z\bar{z} equals [1]a|z|b2\,\mathrm{Re}(z)c|z|^{2}d0
If \omega is a complex cube root of unity, then 1+\omega+\omega^{2} is [1]a1b0c\omegad3
The argument of the complex number -i is [1]a0b\dfrac{\pi}{2}c-\dfrac{\pi}{2}d\pi

Answers to Group A

1. (b) — 10 = 4(2)+2, so i^{10} = i^{2} = -1.

2. (b) — \sqrt{9+16} = 5.

3. (b) — change the sign of the imaginary part only.

4. (c) — (a+ib)(a-ib) = a^{2}+b^{2} = |z|^{2}.

5. (b) — it is the sum of the roots-plus-one identity for z^{3} = 1.

6. (c) — -i is the point (0,-1), a quarter turn clockwise.

Group B — Short answer (5 marks each)

Express \dfrac{2+3i}{1-2i} in the form a+ib and state its conjugate. [5]
Find the modulus and the principal argument of z = -1-i, and represent it on an Argand diagram. [5]
Find real x and y if (x+iy)(2-i) = 7 + i. [5]
Find the square roots of -5+12i. [5]
Show that (2+3i)(2-3i) = |2+3i|^{2}, and state the general property this illustrates. [5]
Find the square roots of 7+24i. [5]

Answers to Group B

1. Multiply by the conjugate 1+2i:

\frac{2+3i}{1-2i}\times\frac{1+2i}{1+2i} = \frac{2+4i+3i+6i^{2}}{1^{2}+2^{2}} = \frac{2+7i-6}{5} = \frac{-4+7i}{5}

So z = -\dfrac45 + \dfrac75 i, and its conjugate is -\dfrac45 - \dfrac75 i.

2. a = -1, b = -1, so |z| = \sqrt{1+1} = \sqrt{2}. \alpha = \tan^{-1}\left(\dfrac{1}{1}\right) = \tan^{-1}1 = \dfrac{\pi}{4}. The point (-1,-1) is in the third quadrant, so

\arg z = -\left(\pi - \frac{\pi}{4}\right) = -\frac{3\pi}{4} = -135^{\circ}

On the Argand diagram, plot (-1,-1): the vector points down-left at 45^{\circ} below the negative real axis, with length \sqrt{2}.

3. (x+iy)(2-i) = 2x - xi + 2yi - yi^{2} = (2x+y) + i(2y-x). Equating parts,

2x + y = 7, \qquad -x + 2y = 1

From the first, y = 7-2x. Substituting into the second,

-x + 2(7-2x) = 1 \;\Longrightarrow\; -5x + 14 = 1 \;\Longrightarrow\; x = \frac{13}{5}

Then y = 7 - \dfrac{26}{5} = \dfrac{9}{5}.

x = \frac{13}{5}, \qquad y = \frac{9}{5}

Check. \left(\tfrac{13}{5}+\tfrac95 i\right)(2-i) = \tfrac{26}{5} - \tfrac{13}{5}i + \tfrac{18}{5}i - \tfrac95 i^{2} = \tfrac{26+9}{5} + \tfrac{5}{5}i = 7+i. Correct.

4. Let \sqrt{-5+12i} = x+iy. Then x^{2}-y^{2} = -5 and 2xy = 12, while

x^{2}+y^{2} = \sqrt{25+144} = \sqrt{169} = 13

Adding: 2x^{2} = 8 \Rightarrow x = \pm2. Subtracting: 2y^{2} = 18 \Rightarrow y = \pm3. As 2xy = 12 > 0 the signs match, so

\sqrt{-5+12i} = \pm(2+3i) \qquad\left[\text{check } (2+3i)^{2} = 4+12i-9 = -5+12i\right]

5. (2+3i)(2-3i) = 4 - 6i + 6i - 9i^{2} = 4+9 = 13, and |2+3i|^{2} = \left(\sqrt{4+9}\right)^{2} = 13. They are equal.

This illustrates z\bar{z} = |z|^{2}: a complex number times its conjugate is the square of its modulus, always a non-negative real number.

6. Let \sqrt{7+24i} = x+iy. Then x^{2}-y^{2} = 7, 2xy = 24, and

x^{2}+y^{2} = \sqrt{49+576} = \sqrt{625} = 25

Adding: 2x^{2} = 32 \Rightarrow x = \pm4. Subtracting: 2y^{2} = 18 \Rightarrow y = \pm3. Since 2xy = 24 > 0 the signs match:

\sqrt{7+24i} = \pm(4+3i) \qquad\left[\text{check } (4+3i)^{2} = 16+24i-9 = 7+24i\right]

Group C — Long answer (8 marks each)

(a) Define the modulus and the conjugate of a complex number. [2](b) Prove that z\bar{z} = |z|^{2} and that |z_1z_2| = |z_1||z_2|. [4](c) Hence find \dfrac{1}{3-2i} in the form a+ib. [2]
(a) Represent z = 3+2i on an Argand diagram and find |z| and \arg z. [4](b) Express z = -1+i\sqrt{3} in polar form and mark it, together with \bar{z}, on the same diagram. [4]
(a) Show that the cube roots of unity are 1, \omega, \omega^{2}, where \omega = \dfrac{-1+i\sqrt{3}}{2}, and prove 1+\omega+\omega^{2} = 0. [4](b) Find the square roots of -8-6i. [4]

Answers to Group C

1. (a) For z = a+ib (a, b real), the modulus is |z| = \sqrt{a^{2}+b^{2}}, the distance of the point (a,b) from the origin; the conjugate is \bar{z} = a-ib, the reflection of that point in the real axis.

(b) z\bar{z} = (a+ib)(a-ib) = a^{2} - i^{2}b^{2} = a^{2}+b^{2} = |z|^{2}.

For the product, let z_1 = a+ib and z_2 = c+id. Then z_1z_2 = (ac-bd) + i(ad+bc), so

|z_1z_2|^{2} = (ac-bd)^{2} + (ad+bc)^{2} = a^{2}c^{2}+b^{2}d^{2}+a^{2}d^{2}+b^{2}c^{2}

= \left(a^{2}+b^{2}\right)\left(c^{2}+d^{2}\right) = |z_1|^{2}|z_2|^{2}

Taking non-negative square roots, |z_1z_2| = |z_1||z_2|.

\frac{1}{3-2i} = \frac{3+2i}{9+4} = \frac{3}{13} + \frac{2}{13}\,i

2. (a) z = 3+2i is the point (3,2), in the first quadrant.

|z| = \sqrt{9+4} = \sqrt{13} \approx 3.606

\arg z = \tan^{-1}\frac{2}{3} \approx 33.69^{\circ} \approx 0.588 \text{ rad}

(b) z = -1+i\sqrt{3}: |z| = \sqrt{1+3} = 2 and, since (-1,\sqrt{3}) is in the second quadrant, \arg z = \pi - \dfrac{\pi}{3} = \dfrac{2\pi}{3}. Hence

z = 2\left(\cos\frac{2\pi}{3} + i\sin\frac{2\pi}{3}\right)

\bar{z} = -1-i\sqrt{3} is the reflection in the real axis: same modulus 2, and argument -\dfrac{2\pi}{3}.

3. (a) z^{3} = 1 \Rightarrow z^{3}-1 = 0 \Rightarrow (z-1)\left(z^{2}+z+1\right) = 0. So either z = 1, or

z = \frac{-1 \pm \sqrt{1-4}}{2} = \frac{-1 \pm i\sqrt{3}}{2}

Writing \omega = \dfrac{-1+i\sqrt{3}}{2}, a direct squaring gives \omega^{2} = \dfrac{-1-i\sqrt{3}}{2}, which is the other root. So the three cube roots are 1, \omega, \omega^{2}.

Adding them: 1 + \dfrac{-1+i\sqrt{3}}{2} + \dfrac{-1-i\sqrt{3}}{2} = 1 + \dfrac{-2}{2} = 0. (Equivalently, \omega and \omega^{2} are the roots of z^{2}+z+1 = 0, whose sum is -1.)

(b) Let \sqrt{-8-6i} = x+iy. Then x^{2}-y^{2} = -8, 2xy = -6, and

x^{2}+y^{2} = \sqrt{64+36} = \sqrt{100} = 10

Adding: 2x^{2} = 2 \Rightarrow x = \pm1. Subtracting: 2y^{2} = 18 \Rightarrow y = \pm3. Since 2xy = -6 < 0 the signs are opposite:

\sqrt{-8-6i} = \pm(1-3i)

Check. (1-3i)^{2} = 1 - 6i + 9i^{2} = 1 - 6i - 9 = -8-6i. Correct.

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