ALGEBRA · HL
Complex Numbers
When $x^2 = -1$.
Section 1 of 10
Introducing $i$
On the real number line, $x^2 = 4$ has solutions $x = 2$ and $x = -2$. Two real answers.
Now try $x^2 = -1$. No real number squared gives a negative.
So $x$ has to live somewhere off the real line — somewhere new.
Call this new number $i$. It's defined so that $i^2 = -1$.
Must learn
1.$i = \sqrt{-1}$
2.$i^2 = -1$
Watch how this lets us simplify roots of negatives using surds rules: $\sqrt{ab} = \sqrt{a}\sqrt{b}$.
$\sqrt{-16} = \sqrt{16} \cdot \sqrt{-1} = 4i$
YOU TRY · 1
Evaluate $i^2$, $i^3$, $i^4$.
$i \cdot i = ?$ Then keep multiplying by $i$.
$i^2 = -1$
$i^3 = i^2 \cdot i = -1 \cdot i = -i$
$i^4 = i^2 \cdot i^2 = (-1)(-1) = 1$
$-1, \; -i, \; 1$
YOU TRY · 2
Simplify $\sqrt{-25}$.
Split into $\sqrt{25} \cdot \sqrt{-1}$.
$\sqrt{-25} = \sqrt{25} \cdot \sqrt{-1}$
$= 5i$
$5i$
Section 2 of 10
Complex roots of quadratics
When $b^2 - 4ac < 0$, the quadratic formula gives the square root of a negative — so the roots are complex, not real.
Worked example
Solve $x^2 - 2x + 5 = 0$.
$a = 1, \quad b = -2, \quad c = 5$
$x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
$= \dfrac{2 \pm \sqrt{4 - 20}}{2}$
$= \dfrac{2 \pm \sqrt{-16}}{2}$
$\sqrt{-16} = \sqrt{16} \cdot \sqrt{-1} = 4i$
$= \dfrac{2 \pm 4i}{2}$
$z = 1 + 2i \quad \text{and} \quad \bar{z} = 1 - 2i$
The two roots come as a conjugate pair — same real part, opposite imaginary parts.
If we draw $y = x^2 - 2x + 5 = (x-1)^2 + 4$, the parabola has its minimum at $(1, 4)$ and never touches the $x$-axis. No real roots — but we still get two complex ones.
Must learn
1.$b^2 - 4ac < 0 \;\Rightarrow\;$ complex roots
2.Complex roots always come in conjugate pairs.
YOU TRY · 1
Solve $x^2 - 4x + 13 = 0$.
Quadratic formula. Watch the sign under the surd.
$a=1, \; b=-4, \; c=13$
$x = \dfrac{4 \pm \sqrt{16 - 52}}{2} = \dfrac{4 \pm \sqrt{-36}}{2}$
$= \dfrac{4 \pm 6i}{2} = 2 \pm 3i$
$x = 2 + 3i$ or $x = 2 - 3i$
YOU TRY · 2
Solve $x^2 + 6x + 25 = 0$.
$b^2 - 4ac = 36 - 100 = -64$.
$x = \dfrac{-6 \pm \sqrt{36 - 100}}{2} = \dfrac{-6 \pm \sqrt{-64}}{2}$
$= \dfrac{-6 \pm 8i}{2} = -3 \pm 4i$
$x = -3 + 4i$ or $x = -3 - 4i$
Section 3 of 10
The form $z = x + yi$
A complex number is written in the form:
$z = x + yi$
where $x$ and $y$ are real numbers, and $i = \sqrt{-1}$.
$x = $ Real part of $z = \operatorname{Re}(z)$
$y = $ Imaginary part of $z = \operatorname{Im}(z)$
The imaginary part is the real number multiplying $i$ — it's $y$, not $yi$.
Quick example
For $z = 3 - 5i$:
$\operatorname{Re}(z) = 3$
$\operatorname{Im}(z) = -5$
YOU TRY · 1
For $z = 4 - 7i$, write down $\operatorname{Re}(z)$ and $\operatorname{Im}(z)$.
Match it to $z = x + yi$.
$z = 4 + (-7)i$
$\operatorname{Re}(z) = 4, \quad \operatorname{Im}(z) = -7$
$\operatorname{Re}(z) = 4, \quad \operatorname{Im}(z) = -7$
YOU TRY · 2
For $z = -3 + 2i$, write down $\operatorname{Re}(z)$ and $\operatorname{Im}(z)$.
Watch the sign on the real part.
$\operatorname{Re}(z) = -3, \quad \operatorname{Im}(z) = 2$
$\operatorname{Re}(z) = -3, \quad \operatorname{Im}(z) = 2$
Section 4 of 10
The Argand diagram
The Argand diagram is a 2-D plane: the horizontal axis is the Real axis, the vertical axis is the Imaginary axis.
$z = x + yi$ is plotted at the point $(x, y)$.
Worked example — plot and transform
Plot $z = 2 + i$ on an Argand diagram. Then plot:
(i) $2z$ (ii) $iz$
$2z = 2(2 + i) = 4 + 2i$
$iz = i(2 + i) = 2i + i^2 = -1 + 2i$
Using $i^2 = -1$.
Notice: $2z$ stretches $z$ outward along the line from the origin. That's a dilation.
And $iz$ rotates $z$ by $90°$ anti-clockwise about the origin.
Must learn — geometric meaning
1.$kz = kx + kyi$ is a dilation along line $OZ$ where $O = (0,0)$.
2.$k > 1 \;\Rightarrow\;$ moved out from origin.
3.$k < 1 \;\Rightarrow\;$ moved in toward origin.
4.$iz = $ a $90°$ anti-clockwise rotation about $(0,0)$.
YOU TRY · 1
$z = 3 + 2i$. Find $2z$ and $iz$, and state where each lies on the Argand diagram.
$iz$: multiply out and use $i^2 = -1$.
$2z = 2(3+2i) = 6 + 4i$ at $(6, 4)$
$iz = i(3+2i) = 3i + 2i^2 = -2 + 3i$ at $(-2, 3)$
$2z = 6 + 4i$, $iz = -2 + 3i$
Section 5 of 10
Add and subtract
Add real parts to real parts, imaginary parts to imaginary parts. Like collecting like terms.
(i) Worked example
$z = 2 + 3i$, $w = 5 - 4i$. Find:
(i) $z + w$ (ii) $3z + 5w$
$z + w = (2 + 3i) + (5 - 4i)$
$= 7 - i$
$3z + 5w = 3(2 + 3i) + 5(5 - 4i)$
$= 6 + 9i + 25 - 20i$
$= 31 - 11i$
(ii) The parallelogram law
$z = 3 - i$ and $w = -2 - 2i$. Show $z$, $w$, and $z + w$ on an Argand diagram.
$z + w = (3 - i) + (-2 - 2i) = 1 - 3i$
Must learn
1.$O$, $z$, $w$ and $z + w$ form a parallelogram.
YOU TRY · 1
$z = 4 + i$, $w = 2 - 3i$. Find $z + w$ and $2z - w$.
Reals with reals, imaginaries with imaginaries.
$z + w = (4+i) + (2-3i) = 6 - 2i$
$2z - w = 2(4+i) - (2-3i) = 8 + 2i - 2 + 3i = 6 + 5i$
$z + w = 6 - 2i, \quad 2z - w = 6 + 5i$
Section 6 of 10
Multiply
Expand the brackets like ordinary algebra. Then replace $i^2$ with $-1$.
(i) Worked example
$z = 5 + i$, $w = 3 - 2i$. Find $zw$.
$zw = (5 + i)(3 - 2i)$
$= 15 - 10i + 3i - 2i^2$
$i^2 = -1$
$= 15 - 7i + 2$
$= 17 - 7i$
(ii) Conjugate pair product
Simplify $(5 + 3i)(5 - 3i)$.
$= 25 - 9i^2$
$i^2 = -1$
$= 25 + 9$
$= 34$
The middle terms cancelled. The answer is real — that's no accident.
YOU TRY · 1
$z = 3 + 2i$, $w = 4 - i$. Find $zw$.
Multiply out, then collect.
$(3+2i)(4-i) = 12 - 3i + 8i - 2i^2$
$= 12 + 5i + 2$
$= 14 + 5i$
$zw = 14 + 5i$
YOU TRY · 2
Simplify $(1 + i)^2$.
$(a+b)^2 = a^2 + 2ab + b^2$.
$(1+i)^2 = 1 + 2i + i^2$
$= 1 + 2i - 1 = 2i$
$(1+i)^2 = 2i$
Section 7 of 10
The complex conjugate
Flip the sign on the imaginary part. That's the conjugate.
Definition
1.$z = x + yi$ has complex conjugate $\bar{z} = x - yi$.
2.$z \cdot \bar{z} = k$ where $k \in \mathbb{R}$.
Quick check: $(x + yi)(x - yi) = x^2 - y^2 i^2 = x^2 + y^2$. Real.
Worked example
$z = 5 - 6i$. Find:
(i) $z + \bar{z}$ (ii) $z \cdot \bar{z}$
$z = 5 - 6i \quad \Rightarrow \quad \bar{z} = 5 + 6i$
$z + \bar{z} = (5 - 6i) + (5 + 6i)$
$= 10$
$z \cdot \bar{z} = (5 - 6i)(5 + 6i) = 25 - 36i^2$
$= 61$
Both real. The imaginary parts cancelled on addition and on multiplication.
Conjugate Root Theorem
1.If $z$ is a root of $az^2 + bz + c = 0$ where $a, b, c \in \mathbb{R}$, then $\bar{z}$ is also a root.
YOU TRY · 1
$z = 4 - 3i$. Find $z + \bar{z}$ and $z \cdot \bar{z}$.
$\bar{z} = 4 + 3i$.
$z + \bar{z} = (4-3i) + (4+3i) = 8$
$z \cdot \bar{z} = (4-3i)(4+3i) = 16 - 9i^2 = 16 + 9 = 25$
$z + \bar{z} = 8, \quad z \cdot \bar{z} = 25$
YOU TRY · 2
One root of $z^2 - 4z + 13 = 0$ is $2 + 3i$. Write down the other root.
Conjugate Root Theorem. No calculation needed.
Coefficients $1, -4, 13$ are all real.
So the other root is the conjugate: $2 - 3i$.
$2 - 3i$
Section 8 of 10
Division
The trick is the same idea as rationalising a surd denominator — multiply top and bottom by something that clears the bottom.
Compare with surds: $\dfrac{2+\sqrt{3}}{\sqrt{3}} \cdot \dfrac{\sqrt{3}}{\sqrt{3}}$. Same idea.
(i) Dividing by $i$
Simplify $\dfrac{2 + 3i}{i}$.
Multiply above and below by $i$.
$\dfrac{2 + 3i}{i} \cdot \dfrac{i}{i}$
$= \dfrac{2i + 3i^2}{i^2}$
$i \cdot i = i^2 = -1$
$= \dfrac{-3 + 2i}{-1}$
$= 3 - 2i$
(ii) Dividing by $a + bi$
Simplify $\dfrac{3 + 5i}{2 - 6i}$.
Must learn — the move
1.Multiply above and below by the conjugate of the bottom.
$\dfrac{3 + 5i}{2 - 6i} \cdot \dfrac{2 + 6i}{2 + 6i}$
Top:
$(3 + 5i)(2 + 6i) = 6 + 18i + 10i + 30i^2$
$= 6 + 28i - 30 = -24 + 28i$
Bottom:
$(2 - 6i)(2 + 6i) = 4 - 36i^2 = 4 + 36 = 40$
$= \dfrac{-24 + 28i}{40}$
$= \dfrac{-6 + 7i}{10}$
(iii) $\dfrac{z}{\bar{z}}$ in form $a + bi$
$z = 3 + 5i$. Simplify $\dfrac{z}{\bar{z}}$, leaving answer in the form $a + bi$.
$z = 3 + 5i, \quad \bar{z} = 3 - 5i$
$\dfrac{3 + 5i}{3 - 5i} \cdot \dfrac{3 + 5i}{3 + 5i}$
Top:
$(3 + 5i)(3 + 5i) = 9 + 15i + 15i + 25i^2$
$= 9 + 30i - 25 = -16 + 30i$
Bottom:
$(3 - 5i)(3 + 5i) = 9 - 25i^2 = 9 + 25 = 34$
$= \dfrac{-16 + 30i}{34}$
$= \dfrac{-8}{17} + \dfrac{15}{17}i$
YOU TRY · 1
Simplify $\dfrac{1 + i}{2 - i}$. Give answer in the form $a + bi$.
Conjugate of bottom is $2 + i$. Multiply above and below.
$\dfrac{1+i}{2-i} \cdot \dfrac{2+i}{2+i}$
Top: $(1+i)(2+i) = 2 + i + 2i + i^2 = 1 + 3i$
Bottom: $(2-i)(2+i) = 4 - i^2 = 5$
$= \dfrac{1 + 3i}{5} = \dfrac{1}{5} + \dfrac{3}{5}i$
$\dfrac{1}{5} + \dfrac{3}{5}i$
Section 9 of 10
Equality of complex numbers
Two complex numbers are equal when their real parts match and their imaginary parts match. That gives two simultaneous equations from one complex equation.
Must learn
1.Real $=$ Real
2.Imaginary $=$ Imaginary
(i) Worked example — real coefficients
$p(2 + i) + q(3 + 5i) = 7 + 11i$. Find $p$ and $q$.
$2p + pi + 3q + 5qi = 7 + 11i$
Group reals and imaginaries:
$(2p + 3q) + (p + 5q)i = 7 + 11i$
Real: $\;\; 2p + 3q = 7$
Imag: $\;\; p + 5q = 11$
Solve simultaneously. From the second, $p = 11 - 5q$. Sub in first:
$2(11 - 5q) + 3q = 7$
$22 - 10q + 3q = 7$
$-7q = -15 \;\Rightarrow\; q = \dfrac{15}{7}$
$p = 11 - 5 \cdot \dfrac{15}{7} = \dfrac{77 - 75}{7} = \dfrac{2}{7}$
$p = \dfrac{2}{7}, \quad q = \dfrac{15}{7}$
(ii) Worked example — $z$ and $\bar{z}$
$z = 2 + 3i$. Given $pz + q\bar{z} = 7 - 9i$, find $p$ and $q$.
$z = 2 + 3i, \quad \bar{z} = 2 - 3i$
$p(2 + 3i) + q(2 - 3i) = 7 - 9i$
$2p + 3pi + 2q - 3qi = 7 - 9i$
Real: $\;\; 2p + 2q = 7$
Imag: $\;\; 3p - 3q = -9 \;\Rightarrow\; p - q = -3$
Add the equations after halving the first: $p + q = 3.5$ and $p - q = -3$.
$2p = 0.5 \;\Rightarrow\; p = 0.25$
$q = 3.5 - 0.25 = 3.25$
$p = \dfrac{1}{4}, \quad q = \dfrac{13}{4}$
(iii) LC-style — two values of $w$
$w$ is a complex number such that $w\bar{w} - 2iw = 7 - 4i$. Find the two possible values of $w$ in the form $p + qi$ where $p, q \in \mathbb{R}$.
Let $w = p + qi, \quad \bar{w} = p - qi$.
$(p + qi)(p - qi) - 2i(p + qi) = 7 - 4i$
$i^2 = -1$
$p^2 - q^2 i^2 - 2pi - 2qi^2 = 7 - 4i$
$p^2 + q^2 - 2pi + 2q = 7 - 4i$
Real: $\;\; p^2 + q^2 + 2q = 7$
Imag: $\;\; -2p = -4 \;\Rightarrow\; p = 2$
Sub $p = 2$: $\quad 4 + q^2 + 2q = 7$
$q^2 + 2q - 3 = 0$
$(q + 3)(q - 1) = 0$
$q = -3 \quad$ or $\quad q = 1$
$w = 2 - 3i \quad$ or $\quad w = 2 + i$
(iv) "Is real" condition
Given $z = 2 - i\sqrt{3}$, find the real number $t$ such that $z^2 + tz$ is real.
$z^2 + tz = (2 - \sqrt{3}\,i)^2 + t(2 - \sqrt{3}\,i)$
$= 4 - 4\sqrt{3}\,i + 3i^2 + 2t - \sqrt{3}\,ti$
$3i^2 = -3$
$= (4 - 3 + 2t) + (-4\sqrt{3} - \sqrt{3}\,t)i$
$= (1 + 2t) - \sqrt{3}\,(4 + t)i$
Is real
1.Is Real $\;\Rightarrow\;$ Imaginary part $= 0$.
$-\sqrt{3}\,(4 + t) = 0$
$4 + t = 0$
$t = -4$
YOU TRY · 1
$p(1 + i) + q(2 - i) = 5 + i$. Find $p$ and $q$.
Expand, then equate reals and imaginaries.
$p + pi + 2q - qi = 5 + i$
Real: $\;\; p + 2q = 5$
Imag: $\;\; p - q = 1$
Subtract: $\;\; 3q = 4 \;\Rightarrow\; q = \dfrac{4}{3}$
$p = 1 + q = \dfrac{7}{3}$
$p = \dfrac{7}{3}, \quad q = \dfrac{4}{3}$
YOU TRY · 2
Given $z = 3 - i\sqrt{2}$, find the real number $t$ such that $z^2 + tz$ is real.
Same as the worked example. Imaginary part must be zero.
$(3 - \sqrt{2}\,i)^2 = 9 - 6\sqrt{2}\,i + 2i^2 = 7 - 6\sqrt{2}\,i$
$z^2 + tz = (7 + 3t) + (-6\sqrt{2} - \sqrt{2}\,t)i$
Imag $= 0:\;\; -\sqrt{2}\,(6 + t) = 0$
$t = -6$
$t = -6$
Section 10 of 10
Modulus
The modulus of $z$ is its distance from the origin on the Argand diagram. Pythagoras gives the formula.
Must learn
1.$r = |z| = \sqrt{x^2 + y^2}$
$r^2 = x^2 + y^2$
(i) Worked example
$z = 2 + 5i$. Find $|z|$.
Common error first — then the fix.
$|z| = \sqrt{2^2 + (5i)^2} = \sqrt{4 + 25i^2} = \sqrt{-21}$ ✗
You do not square the $i$ along with the $5$. The $y$ in the formula is just the real number $5$.
$|z| = \sqrt{x^2 + y^2} = \sqrt{2^2 + 5^2}$
$|z| = \sqrt{29}$
(ii) Worked example — modulus of an expression
Given $z = 2 - i$, calculate $|z^2 - z + 3|$ where $i^2 = -1$.
$z^2 - z + 3 = (2 - i)^2 - (2 - i) + 3$
$= 4 - 4i + i^2 - 2 + i + 3$
$i^2 = -1$
$= (4 - 1 - 2 + 3) + (-4i + i)$
$= 4 - 3i$
$|4 - 3i| = \sqrt{4^2 + (-3)^2} = \sqrt{16 + 9} = \sqrt{25}$
$|z^2 - z + 3| = 5$
(iii) Prove $|w||z| = |wz|$
Let $w = p + qi$ and $z = x + yi$.
LHS:
$|w| = \sqrt{p^2 + q^2}, \quad |z| = \sqrt{x^2 + y^2}$
$|w||z| = \sqrt{p^2 + q^2} \cdot \sqrt{x^2 + y^2}$
$\sqrt{a}\sqrt{b} = \sqrt{ab}$
$= \sqrt{(p^2 + q^2)(x^2 + y^2)}$
$= \sqrt{p^2 x^2 + p^2 y^2 + q^2 x^2 + q^2 y^2}$
RHS:
$wz = (p + qi)(x + yi) = px + pyi + qxi + qyi^2$
$= (px - qy) + (py + qx)i$
$|wz| = \sqrt{(px - qy)^2 + (py + qx)^2}$
$= \sqrt{p^2 x^2 - 2pqxy + q^2 y^2 + p^2 y^2 + 2pqxy + q^2 x^2}$
The $\pm 2pqxy$ terms cancel.
$= \sqrt{p^2 x^2 + q^2 y^2 + p^2 y^2 + q^2 x^2}$
LHS $=$ RHS ∴ $|w||z| = |wz|$ Q.E.D.
YOU TRY · 1
$z = 3 + 4i$. Find $|z|$.
$|z| = \sqrt{x^2 + y^2}$. Famous triple.
$|z| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25}$
$= 5$
$|z| = 5$
YOU TRY · 2
$z = 1 - i$. Find $|z^2 + 2z|$.
Get $z^2 + 2z$ in form $a + bi$ first. Then modulus.
$z^2 = (1-i)^2 = 1 - 2i + i^2 = -2i$
$2z = 2 - 2i$
$z^2 + 2z = -2i + 2 - 2i = 2 - 4i$
$|2 - 4i| = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5}$
$|z^2 + 2z| = 2\sqrt{5}$
SUM
The lot in one box
Complex numbers toolkit
1.$i = \sqrt{-1}, \;\; i^2 = -1$
2.$z = x + yi$ ($x = \operatorname{Re}(z), \; y = \operatorname{Im}(z)$)
3.$b^2 - 4ac < 0 \;\Rightarrow\;$ complex roots in conjugate pairs.
4.Argand: $z$ at $(x, y)$. $kz$ dilates, $iz$ rotates $90°$.
5.$O, z, w, z+w$ form a parallelogram.
6.Conjugate $\bar{z} = x - yi$. $z + \bar{z}, \; z\bar{z}$ both real.
7.Conjugate Root Theorem: real-coefficient quadratics give roots in conjugate pairs.
8.Division: multiply above and below by the conjugate of the bottom.
9.Equality: Real $=$ Real, Imag $=$ Imag.
10.$|z| = \sqrt{x^2 + y^2}$. $|wz| = |w||z|$.
End of lesson
Complex Numbers — HL · Mathslive.ie