Core Pure · Complex Numbers · Year 1

Imaginary & Complex Numbers

A single new number, i, opens up a whole second dimension of numbers where every polynomial finally has as many roots as its degree.

14 min read 5 worked examples

Extending the number system

Definition 1.1 — the imaginary unit

The equation $x^2=-1$ has no solution among the real numbers, since a square can never be negative. We define a new number $i$ satisfying $$i^2=-1$$ Every other power of $i$ follows from this: $i^3=i^2\cdot i=-i$, $i^4=(i^2)^2=1$, and the cycle $i,-1,-i,1$ repeats every four powers.

Definition 1.2 — complex numbers

A complex number is any number of the form $z=a+bi$, where $a,b\in\mathbb{R}$. The real part is $\operatorname{Re}(z)=a$ and the imaginary part is $\operatorname{Im}(z)=b$ (note: $\operatorname{Im}(z)$ is the real number $b$, not $bi$). The set of all complex numbers is denoted $\mathbb{C}$. If $b=0$, $z$ is real; if $a=0$ and $b\neq0$, $z$ is called purely imaginary.

Writing square roots of negative numbers in terms of i

Method 1.3 — simplifying √(negative number)

Any square root of a negative real number can be rewritten using $i$: split the number under the root into a positive factor and $-1$, then pull $\sqrt{-1}=i$ out separately. This is exactly the same surd-manipulation technique from Pure Year 1, just applied to a negative radicand instead of a positive one.

$$\sqrt{-n}=\sqrt{n\times(-1)}=\sqrt{n}\sqrt{-1}=i\sqrt{n}, \qquad n>0$$

Example A · Writing roots of negative numbers in terms of i

Write each of the following in the form $bi$: (a) $\sqrt{-36}$, (b) $\sqrt{-28}$.

(a) $\sqrt{-36}=\sqrt{36\times(-1)}=\sqrt{36}\sqrt{-1}=6i$.

(b) $\sqrt{-28}=\sqrt{28\times(-1)}=\sqrt{28}\sqrt{-1}=\sqrt{4}\sqrt{7}\sqrt{-1}=2\sqrt{7}\,i$.

Watch out — where to place the i

Write $2\sqrt7\,i$ (or equivalently $2i\sqrt7$) rather than $2\sqrt{7i}$ — the second version puts $i$ inside the root, which means something completely different and is a common exam-losing slip. Keep $i$ clearly separated from any surd it's multiplying.

Arithmetic with complex numbers

Method — addition, subtraction, and real scalar multiplication

Complex numbers combine like algebraic binomials in $i$: add or subtract real parts together and imaginary parts together, treating $i$ as an unknown (but always simplifying using $i^2=-1$ once a product arises).

$$(a+bi)\pm(c+di)=(a\pm c)+(b\pm d)i,$$ $$k(a+bi)=ka+kbi \text{ for } k\in\mathbb{R}$$

Example 1 · Simplifying powers of i

Simplify (a) $i^5$, (b) $i^{11}$, (c) $i^{100}$.

(a) $i^5=i^4\cdot i=1\cdot i=i$.

(b) $i^{11}=i^8\cdot i^3=(i^4)^2\cdot i^3=1\cdot(-i)=-i$.

Example 2 · Adding and subtracting complex numbers

Given $z_1=5-3i$ and $z_2=-2+7i$, find (a) $z_1+z_2$, (b) $z_1-z_2$, (c) $3z_1-2z_2$.

(a) $z_1+z_2=(5-2)+(-3+7)i=3+4i$.

(b) $z_1-z_2=(5-(-2))+(-3-7)i=7-10i$.

Example 3 · Solving a simple complex equation

Find the real numbers $x$ and $y$ such that $(x+2y)+(2x-y)i=7+i$.

Equating real and imaginary parts (this is only valid because $x,y$ are real, so both sides are already written in standard $a+bi$ form):

$x+2y=7$ and $2x-y=1$.

From the second equation, $y=2x-1$. Substituting: $x+2(2x-1)=7 \Rightarrow x+4x-2=7 \Rightarrow 5x=9 \Rightarrow x=\tfrac95$.

Then $y=2(\tfrac95)-1=\tfrac{18}{5}-1=\tfrac{13}{5}$.

Exam technique — equating real and imaginary parts

Whenever an equation between two complex numbers is given in the form $a+bi=c+di$ with $a,b,c,d$ real, you may equate separately: $a=c$ and $b=d$. This is one of the most frequently used techniques in this topic and turns a single complex equation into a pair of simultaneous real equations.

Example 4 · Working with purely imaginary numbers

Given that $z=(k-3)+(2k+1)i$ is purely imaginary, where $k$ is real, find the value of $k$ and state the value of $z$.

Purely imaginary means the real part is zero (and the imaginary part is non-zero): $k-3=0 \Rightarrow k=3$.

Check the imaginary part is non-zero: $2(3)+1=7\neq0$, so this is valid.

$z=0+7i=7i$.

Common mistake

"Purely imaginary" requires the real part to be exactly zero and the imaginary part to be non-zero — the number $0$ itself is usually not classified as purely imaginary. Always check both conditions, not just that the real part vanishes.

Exam-style questions

Question 1 6 marks

(a) Simplify $i^7+i^{18}$, giving your answer in the form $a+bi$. (b) Given $z_1=4+i$ and $z_2=-1-5i$, find $2z_1-3z_2$ in the form $a+bi$.

(a) $i^7=i^4\cdot i^3=1\cdot(-i)=-i$. $i^{18}=i^{16}\cdot i^2=(i^4)^4\cdot(-1)=1\cdot(-1)=-1$. Sum: $-1-i$. (M1, A1, A1)

(b) $2z_1=8+2i$; $3z_2=-3-15i$. $2z_1-3z_2=(8-(-3))+(2-(-15))i=11+17i$. (M1, A1, A1)

Question 2 6 marks

Find the real numbers $p$ and $q$ such that $(2p-q)+(p+3q)i=8-5i$.

Equating real and imaginary parts: $2p-q=8$ and $p+3q=-5$. (M1, A1)

From the first equation, $q=2p-8$. Substituting: $p+3(2p-8)=-5 \Rightarrow p+6p-24=-5 \Rightarrow 7p=19 \Rightarrow p=\tfrac{19}{7}$. (M1, A1)

$q=2(\tfrac{19}{7})-8=\tfrac{38}{7}-\tfrac{56}{7}=-\tfrac{18}{7}$. (A1, A1)

Question 3 5 marks

Given that $z=(3k-1)+(k^2-4)i$ is real, where $k$ is real, find the possible values of $k$ and the corresponding value(s) of $z$.

$z$ real $\Rightarrow$ imaginary part is zero: $k^2-4=0 \Rightarrow k=\pm2$. (M1, A1)

When $k=2$: $z=3(2)-1=5$. When $k=-2$: $z=3(-2)-1=-7$. (M1, A1, A1)

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Multiplying & Dividing Complex Numbers