The Imaginary Unit and Complex Numbers

Complex numbers are the gateway to almost everything in the Further Mathematics compulsory pure content. Once you accept a single new number — the imaginary unit $i$i, whose square is $-1$−1 — every polynomial factorises completely, trigonometric identities fall out of algebra, and a whole geometry of rotation and scaling opens up. This lesson builds that foundation rigorously: what $i$i is, what a complex number is, and the algebra you must own before anything else.

---

Where this sits in AQA 7367

This is the entry point to the complex numbers strand of the compulsory pure content examined on Paper 1 and Paper 2 (each 2 hours, 100 marks, 33⅓% of the A-Level). The wider strand goes on to Argand diagrams, modulus–argument and exponential form, De Moivre's theorem, $n$nth roots and roots of unity, and loci. This first lesson is dominated by AO1 (selecting and carrying out routine procedures — defining $i$i, simplifying powers, equating parts, solving quadratics) with the first taste of AO2 (reasoning: why complex roots of real quadratics pair up). Get the AO1 fluency automatic now and the later AO2/AO3 problem-solving has somewhere to stand.

---

Core theory: building $\mathbb{C}$C from a single new symbol

You already know that some quadratics have no real roots. Consider

$x^2 + 1 = 0 \;\Longrightarrow\; x^2 = -1.$x2+1=0⟹x2=−1.

No real number squares to a negative, so over $\mathbb{R}$R this has no solution. The Further-Maths move is not to give up but to extend the number system: we adjoin a new symbol $i$i with the single defining property

$i^2 = -1.$i2=−1.

It is cleaner to take $i^2 = -1$i2=−1 as the definition and to read "$i = \sqrt{-1}$i=−1​" only as informal shorthand — the radical sign behaves badly on negatives (see Common errors). From this one rule, all the arithmetic follows.

Historically, complex numbers entered mathematics not through quadratics (mathematicians were happy to declare $x^2 = -1$x2=−1 "impossible") but through cubics: the sixteenth-century formula for the real roots of a cubic sometimes required taking square roots of negatives en route, even when all three final roots were real. The "imaginary" intermediate quantities cancelled to give correct real answers, which forced mathematicians to take $i$i seriously. That is a recurring theme: $i$i is a legitimate tool for getting real results, not a mere curiosity. You will see the same phenomenon when De Moivre's theorem produces real trigonometric identities from complex algebra.

A complex number is any number of the form

$z = a + bi, \qquad a, b \in \mathbb{R},$z=a+bi,a,b∈R,

where $a = \operatorname{Re}(z)$a=Re(z) is the real part and $b = \operatorname{Im}(z)$b=Im(z) is the imaginary part. Note carefully that $\operatorname{Im}(z)$Im(z) is the real coefficient of $i$i: for $z = 3 + 2i$z=3+2i, $\operatorname{Im}(z) = 2$Im(z)=2, not $2i$2i.

$z$z	$\operatorname{Re}(z)$Re(z)	$\operatorname{Im}(z)$Im(z)	Type
$3 + 2i$3+2i	$3$3	$2$2	general
$-1 + 4i$−1+4i	$-1$−1	$4$4	general
$5$5	$5$5	$0$0	purely real
$-3i$−3i	$0$0	$-3$−3	purely imaginary

The set of all complex numbers is written $\mathbb{C}$C. Every real number $a$a is the complex number $a + 0i$a+0i, so the reals sit inside the complex numbers, completing the chain of number systems you have built since school:

$\mathbb{N} \subset \mathbb{Z} \subset \mathbb{Q} \subset \mathbb{R} \subset \mathbb{C}.$N⊂Z⊂Q⊂R⊂C.

A number with $b = 0$b=0 is purely real; one with $a = 0$a=0 and $b \neq 0$b=0 is purely imaginary; and $0 = 0 + 0i$0=0+0i is the only number that is both.

Why every quadratic now has roots

The whole point of adjoining $i$i is that quadratics never fail again. For $az^2 + bz + c = 0$az2+bz+c=0 with real $a \neq 0$a=0, completing the square gives the familiar formula

$z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a},$z=2a−b±b2−4ac​​,

and the discriminant $\Delta = b^2 - 4ac$Δ=b2−4ac decides the type of the roots:

Discriminant	Roots over $\mathbb{C}$C
$\Delta > 0$Δ>0	two distinct real roots
$\Delta = 0$Δ=0	one repeated real root
$\Delta < 0$Δ<0	a complex conjugate pair $\dfrac{-b}{2a} \pm \dfrac{\sqrt{4ac - b^2}}{2a}\,i$2a−b​±2a4ac−b2​​i

When $\Delta < 0$Δ<0 we write $\sqrt{\Delta} = \sqrt{-(4ac - b^2)} = i\sqrt{4ac - b^2}$Δ​=−(4ac−b2)​=i4ac−b2​, and the $\pm$± delivers the conjugate pair automatically. So every quadratic with real coefficients has exactly two roots in $\mathbb{C}$C (counting the repeated case as two) — the first hint of the Fundamental Theorem of Algebra discussed later.

Powers of $i$i cycle with period 4

Because $i^2 = -1$i2=−1, higher powers repeat in a cycle of length four:

$i^1 = i, \quad i^2 = -1, \quad i^3 = i^2 \cdot i = -i, \quad i^4 = (i^2)^2 = 1, \quad i^5 = i, \ldots$i1=i,i2=−1,i3=i2⋅i=−i,i4=(i2)2=1,i5=i,…

So for any positive integer $n$n, write $n = 4q + r$n=4q+r with remainder $r \in \{0,1,2,3\}$r∈{0,1,2,3}; then, since $i^{4q} = (i^4)^q = 1^q = 1$i4q=(i4)q=1q=1,

$i^n = i^{4q + r} = i^{4q}\cdot i^r = i^r = \begin{cases} 1 & r = 0 \\ i & r = 1 \\ -1 & r = 2 \\ -i & r = 3. \end{cases}$in=i4q+r=i4q⋅ir=ir=⎩⎨⎧​1i−1−i​r=0r=1r=2r=3.​

The same idea handles negative powers. Because $i^4 = 1$i4=1, the cycle extends both ways: $i^{-1} = \dfrac{1}{i} = \dfrac{1}{i}\cdot\dfrac{i}{i} = \dfrac{i}{i^2} = \dfrac{i}{-1} = -i$i−1=i1​=i1​⋅ii​=i2i​=−1i​=−i, and likewise $i^{-2} = -1$i−2=−1, $i^{-3} = i$i−3=i, $i^{-4} = 1$i−4=1. So you may reduce any integer index modulo $4$4. For instance $i^{-7} = i^{-7 + 8} = i^{1} = i$i−7=i−7+8=i1=i.

A useful consequence worth memorising: any four consecutive powers of $i$i sum to zero, since $i^k(1 + i + i^2 + i^3) = i^k(1 + i - 1 - i) = 0$ik(1+i+i2+i3)=ik(1+i−1−i)=0. This collapses long alternating sums instantly.

Square roots of negative reals

For $k > 0$k>0 we write

$\sqrt{-k} = \sqrt{k}\,i,$−k​=k​i,

so $\sqrt{-9} = 3i$−9​=3i, $\sqrt{-25} = 5i$−25​=5i and $\sqrt{-2} = \sqrt{2}\,i$−2​=2​i. Always place the $i$i outside the surd — $\sqrt{2}\,i$2​i, never the ambiguous $\sqrt{2i}$2i​.

Equality of complex numbers

Two complex numbers are equal exactly when they match in both components:

$a + bi = c + di \iff a = c \;\text{ and }\; b = d.$a+bi=c+di⟺a=c and b=d.

This single statement — comparing (or "equating") real and imaginary parts — turns one complex equation into two real equations and is the most-used technique in the whole topic. The reason it works is that $1$1 and $i$i are "independent directions": no real multiple of $1$1 can ever equal a non-zero real multiple of $i$i (that would put a non-zero real number on the imaginary axis). So if $a + bi = c + di$a+bi=c+di, rearranging gives $(a - c) = (d - b)i$(a−c)=(d−b)i; the left side is real and the right side is imaginary, and the only number that is both is $0$0. Hence $a = c$a=c and $b = d$b=d — two equations from one.

A note on conventions

Three notational habits will save you marks throughout the course:

• Write complex numbers in the standard form $a + bi$a+bi with the real part first; a final answer like $2i + 3$2i+3 should be tidied to $3 + 2i$3+2i.
• Keep the imaginary unit outside surds and fractions where it reads cleanly: prefer $\tfrac{1}{5} + \tfrac{7}{5}i$51​+57​i to $\tfrac{1 + 7i}{5}$51+7i​ when a question asks for the form $a + bi$a+bi.
• Use $z, w$z,w for complex variables and $a, b, c, d, x, y$a,b,c,d,x,y for the real numbers inside them; mixing the two is a common source of confusion in multi-part questions.

---

Worked examples (with mark scheme)

Example 1 — a high power of $i$i

Find $i^{102}$i102.

$102 = 4 \times 25 + 2 \quad (\text{M1: divide by 4, read remainder})$102=4×25+2(M1: divide by 4, read remainder) $i^{102} = i^2 = -1. \quad (\text{A1})$i102=i2=−1.(A1)

(M1 for reducing the index modulo 4; A1 for $-1$−1.)

Example 2 — equating real and imaginary parts

Find real $x, y$x,y with $(x + 2) + (3y - 1)i = 5 + 8i$(x+2)+(3y−1)i=5+8i.

$\text{Real parts:} \quad x + 2 = 5 \;\Rightarrow\; x = 3. \quad (\text{M1 equate real, A1})$Real parts:x+2=5⇒x=3.(M1 equate real, A1) $\text{Imag parts:} \quad 3y - 1 = 8 \;\Rightarrow\; y = 3. \quad (\text{M1 equate imag, A1})$Imag parts:3y−1=8⇒y=3.(M1 equate imag, A1)

(M1 for each correct comparison; A1 for each value. The whole method hinges on comparing components — write the two equations explicitly.)

Example 3 — a quadratic with complex roots

Solve $z^2 + 2z + 5 = 0$z2+2z+5=0.

$\Delta = b^2 - 4ac = 2^2 - 4(1)(5) = 4 - 20 = -16. \quad (\text{M1 discriminant})$Δ=b2−4ac=22−4(1)(5)=4−20=−16.(M1 discriminant) $z = \frac{-2 \pm \sqrt{-16}}{2} = \frac{-2 \pm 4i}{2} = -1 \pm 2i. \quad (\text{M1 surd of negative}\to i;\; \text{A1 both roots})$z=2−2±−16​​=2−2±4i​=−1±2i.(M1 surd of negative→i;A1 both roots)

The roots $-1 + 2i$−1+2i and $-1 - 2i$−1−2i form a conjugate pair — always the case for a real-coefficient quadratic with $\Delta < 0$Δ<0. Check: sum $= -2 = -b/a$=−2=−b/a ✓, product $= (-1)^2 + 2^2 = 5 = c/a$=(−1)2+22=5=c/a ✓.

Example 4 — surds of negatives, done safely

Simplify $\sqrt{-9}\cdot\sqrt{-4}$−9​⋅−4​.

The tempting (and wrong) move is $\sqrt{-9}\cdot\sqrt{-4} = \sqrt{(-9)(-4)} = \sqrt{36} = 6$−9​⋅−4​=(−9)(−4)​=36​=6. The surd-product rule $\sqrt{a}\sqrt{b} = \sqrt{ab}$a​b​=ab​ is invalid for negatives. Instead, convert each surd to $i$i-form first:

$\sqrt{-9}\cdot\sqrt{-4} = (3i)(2i) = 6i^2 = -6. \quad (\text{M1 convert each to } i\text{-form;}\; \text{A1})$−9​⋅−4​=(3i)(2i)=6i2=−6.(M1 convert each to i-form;A1)

(M1 for writing each root as $\sqrt{k}\,i$k​i before multiplying; A1 for $-6$−6. The correct answer is the negative of the naive one — a classic discriminator.)

Example 5 — combining powers of $i$i

Simplify $3i^{5} - 2i^{11} + i^{0}$3i5−2i11+i0.

$i^5 = i,\quad i^{11} = i^{8}\cdot i^{3} = i^3 = -i,\quad i^0 = 1. \quad (\text{M1 reduce each index})$i5=i,i11=i8⋅i3=i3=−i,i0=1.(M1 reduce each index) $3i^5 - 2i^{11} + i^0 = 3i - 2(-i) + 1 = 1 + 5i. \quad (\text{A1})$3i5−2i11+i0=3i−2(−i)+1=1+5i.(A1)

(M1 for reducing every power mod 4; A1 for $1 + 5i$1+5i.)

Example 6 — equating parts to find an unknown complex number

Find the complex number $z = x + yi$z=x+yi (with $x, y$x,y real) satisfying $2z + 3\bar z = 10 - 4i$2z+3zˉ=10−4i, where $\bar z = x - yi$zˉ=x−yi is the conjugate.

Substitute and expand:

$2(x + yi) + 3(x - yi) = (2x + 3x) + (2y - 3y)i = 5x - yi. \quad (\text{M1 substitute, collect})$2(x+yi)+3(x−yi)=(2x+3x)+(2y−3y)i=5x−yi.(M1 substitute, collect)

Now compare with $10 - 4i$10−4i:

$\text{Real:}\; 5x = 10 \Rightarrow x = 2; \qquad \text{Imag:}\; -y = -4 \Rightarrow y = 4. \quad (\text{M1 equate, A1})$Real:5x=10⇒x=2;Imag:−y=−4⇒y=4.(M1 equate, A1)

So $z = 2 + 4i$z=2+4i. (M1 substituting $z$z and $\bar z$zˉ; M1 comparing parts; A1 for $z = 2 + 4i$z=2+4i. Check: $2(2+4i) + 3(2-4i) = (4+8i)+(6-12i) = 10 - 4i$2(2+4i)+3(2−4i)=(4+8i)+(6−12i)=10−4i ✓.) This "let $z = x + yi$z=x+yi, substitute, compare parts" pattern solves a large fraction of all complex-equation questions — it is worth drilling.

---

Specimen-style exam question

(specimen-style — not from any past paper)

The quadratic equation $z^2 + pz + q = 0$z2+pz+q=0, where $p, q$p,q are real, has $z = 3 - 2i$z=3−2i as one root. (a) Write down the other root. (b) Find $p$p and $q$q.

(a) Real coefficients $\Rightarrow$⇒ roots are a conjugate pair, so the other root is $3 + 2i$3+2i.

(b) Using sum and product of roots,

$\text{sum} = (3 - 2i) + (3 + 2i) = 6 = -p \;\Rightarrow\; p = -6,$sum=(3−2i)+(3+2i)=6=−p⇒p=−6, $\text{product} = (3 - 2i)(3 + 2i) = 9 + 4 = 13 = q.$product=(3−2i)(3+2i)=9+4=13=q.

So $z^2 - 6z + 13 = 0$z2−6z+13=0. (Verify: $\Delta = 36 - 52 = -16 < 0$Δ=36−52=−16<0 ✓, consistent with complex roots.)

(c) Hence, or otherwise, solve $z^2 - 6z + 13 = 0$z2−6z+13=0 directly and confirm the roots.

Using the quadratic formula on the equation just found,

$z = \frac{6 \pm \sqrt{36 - 52}}{2} = \frac{6 \pm \sqrt{-16}}{2} = \frac{6 \pm 4i}{2} = 3 \pm 2i,$z=26±36−52​​=26±−16​​=26±4i​=3±2i,

which reproduces the given root $3 - 2i$3−2i and its conjugate $3 + 2i$3+2i. The word "Hence" rewards reusing the equation from (b) rather than starting afresh — a small efficiency that examiners credit.

---

Synoptic links

• A-Level Maths (quadratics, discriminant). Everything here is the discriminant story continued: $\Delta < 0$Δ<0 used to mean "no roots"; now it means "a complex conjugate pair." Sum/product of roots ( $-b/a,\ c/a$−b/a, c/a ) carries straight over.
• Conjugates and polynomial roots (next lesson) generalises the conjugate-pair observation to cubics and quartics.
• Argand diagrams / modulus–argument form turn $a + bi$a+bi into geometry, where multiplication by $i$i becomes a $90^\circ$90∘ rotation.
• Surds and indices underpin the powers-of-$i$i cycle and the handling of $\sqrt{-k}$−k​.
• Proof and the Fundamental Theorem of Algebra: the guarantee that every polynomial factorises over $\mathbb{C}$C (touched on below) sits behind the entire roots-of-polynomials and roots-of-unity material later in the strand.

---

Mark-scheme literacy

• "Solve" expects exact surd/complex form ($-1 \pm 2i$−1±2i), not decimals.
• "Write down" signals one mark for a stated result with minimal working — e.g. quoting the conjugate root. Don't over-work it.
• When you "equate real and imaginary parts," examiners want the two real equations visible; an unsupported final answer can forfeit the method marks even if the values are right.
• Follow-through (ft) marks often rescue a slip: if you mis-compute $\Delta$Δ but then apply the formula correctly, the A1 method credit can still follow through — so always show the formula step.
• "Find" and "Determine" expect a supported answer; a bare value with no working risks losing method marks even when correct. Conversely, "State" or "Write down" signals that minimal working is expected — don't waste time over-justifying.
• When a question gives a value to several parts (e.g. "$z = 2 + i$z=2+i"), it usually intends you to reuse it; recomputing it from scratch in a later part can forfeit a "hence" mark.

---

Grade-band model answers

Question: "Solve $z^2 - 4z + 13 = 0$z2−4z+13=0, giving your answers in the form $a + bi$a+bi."

Mid-band response. $\Delta = 16 - 52 = -36, \qquad z = \frac{4 \pm \sqrt{-36}}{2} = \frac{4 \pm 6i}{2} = 2 \pm 3i.$Δ=16−52=−36,z=24±−36​​=24±6i​=2±3i. Examiner-style commentary: Correct and efficient; full marks for the computation. It states the answers but offers no check, so a single arithmetic slip would go undetected.

Stronger response. As above, plus: "Since the coefficients are real and $\Delta < 0$Δ<0, the roots must be a conjugate pair, consistent with $2 + 3i$2+3i and $2 - 3i$2−3i." Examiner-style commentary: The explicit conjugate-pair reasoning is exactly the AO2 communication examiners reward — the candidate explains why the answer has the form it does.

Top-band response. As Stronger, with a verification: "Sum of roots $= 4 = -b/a$=4=−b/a ✓; product $= 2^2 + 3^2 = 13 = c/a$=22+32=13=c/a ✓, confirming the solution." Examiner-style commentary: Self-checking against the coefficients shows genuine command of the structure of quadratics. This is the habit that prevents lost accuracy marks across a whole paper.

---

Common misconceptions

1. "$i$i is just $\sqrt{-1}$−1​ and obeys all surd rules." It is safer to treat $i^2 = -1$i2=−1 as the definition. The naive rule $\sqrt{a}\sqrt{b} = \sqrt{ab}$a​b​=ab​ fails for negatives (see Common errors).
2. "The imaginary part of $3 + 2i$3+2i is $2i$2i." It is the real number $2$2. $\operatorname{Im}(z)$Im(z) is a real number.
3. "A negative discriminant means no solutions." True only over $\mathbb{R}$R. Over $\mathbb{C}$C it gives a complex conjugate pair.
4. "$i^{102}$i102 needs a big calculation." Only the index mod 4 matters: $i^{102} = i^2 = -1$i102=i2=−1.
5. "Complex numbers can be ordered like reals." There is no order on $\mathbb{C}$C compatible with arithmetic — "$i > 0$i>0" and "$i < 0$i<0" both lead to contradictions, so inequalities such as $z_1 < z_2$z1​<z2​ are meaningless for non-real numbers.
6. "$0$0 is purely real but not imaginary." $0 = 0 + 0i$0=0+0i is the unique number that is both purely real and purely imaginary.
7. "$\sqrt{-9} = \pm 3i$−9​=±3i." As a quadratic root you do write $\pm$±, but the principal square-root notation $\sqrt{-9}$−9​ denotes the single value $3i$3i.
8. "$i$i is some huge or infinitesimal number." It is neither large nor small — it simply is not on the real line at all. Asking whether $i$i is "big" is like asking whether "north" is bigger than "up": it is a different direction.
9. "A complex equation gives one equation." Comparing parts always yields two real equations (one real, one imaginary), so a single complex condition typically pins down two real unknowns.

---

Common errors

• Mishandling $\sqrt{-1}\sqrt{-1}$−1​−1​. Writing $\sqrt{-1}\sqrt{-1} = \sqrt{(-1)(-1)} = \sqrt{1} = 1$−1​−1​=(−1)(−1)​=1​=1 is wrong: the surd-multiplication rule is invalid for negatives. Correctly, $\sqrt{-1}\sqrt{-1} = i \cdot i = i^2 = -1$−1​−1​=i⋅i=i2=−1.
• Dropping the factor of 2 when simplifying $\dfrac{-2 \pm 4i}{2}$2−2±4i​: the answer is $-1 \pm 2i$−1±2i, not $-2 \pm 2i$−2±2i — divide both terms.
• Sign slip on $i^3$i3: $i^3 = -i$i3=−i, not $i$i.
• Forgetting $i^2 = -1$i2=−1 mid-expansion, leaving a stray $i^2$i2 in the final answer.
• Writing $\sqrt{2i}$2i​ when you mean $\sqrt{2}\,i$2​i — ambiguous and easy to misread.
• Combining surds of negatives via $\sqrt{a}\sqrt{b} = \sqrt{ab}$a​b​=ab​ — invalid here; convert each to $i$i-form first (Example 4).
• Leaving $i^2$i2, $i^3$i3 or $i^4$i4 unsimplified in a final answer instead of reducing to $-1$−1, $-i$−i, $1$1.

---

Going further (stretch)

A natural question: why is one new symbol enough? The Fundamental Theorem of Algebra answers it — every non-constant polynomial with complex coefficients has a root in $\mathbb{C}$C. So having added $i$i to solve $x^2 + 1 = 0$x2+1=0, we never need to extend again: $\mathbb{C}$C is algebraically closed. Adjoining $i$i is exactly forming $\mathbb{R}[x]/(x^2+1)$R[x]/(x2+1) — the reals with a symbol $x$x satisfying $x^2 + 1 = 0$x2+1=0 — which is how university algebra constructs $\mathbb{C}$C rigorously, sidestepping any worry about "$\sqrt{-1}$−1​."

A STEP-flavoured taster: find all $z = a + bi$z=a+bi (with $a, b$a,b real) such that $z^2 = i$z2=i. Comparing parts in $(a + bi)^2 = a^2 - b^2 + 2abi = 0 + 1\cdot i$(a+bi)2=a2−b2+2abi=0+1⋅i gives $a^2 - b^2 = 0$a2−b2=0 and $2ab = 1$2ab=1. From the first $a = \pm b$a=±b; the second needs $ab > 0$ab>0, forcing $a = b$a=b with $2a^2 = 1$2a2=1, so $a = b = \tfrac{1}{\sqrt 2}$a=b=2​1​. Hence $z = \pm\tfrac{1}{\sqrt 2}(1 + i)$z=±2​1​(1+i) — the two square roots of $i$i, found with nothing beyond equating parts.

This "equate parts" method is in fact a complete algorithm for square roots of any complex number $w = p + qi$w=p+qi: set $(a + bi)^2 = p + qi$(a+bi)2=p+qi, so $a^2 - b^2 = p$a2−b2=p and $2ab = q$2ab=q; square and add to get $(a^2 + b^2)^2 = (a^2 - b^2)^2 + (2ab)^2 = p^2 + q^2$(a2+b2)2=(a2−b2)2+(2ab)2=p2+q2, hence $a^2 + b^2 = \sqrt{p^2 + q^2}$a2+b2=p2+q2​. Combining this with $a^2 - b^2 = p$a2−b2=p gives $a^2$a2 and $b^2$b2 directly, and the sign of $q = 2ab$q=2ab fixes whether $a, b$a,b share a sign. You will meet exactly this technique again, dressed up as De Moivre's $n$nth-root formula, later in the course — but it is reassuring to know it follows from nothing more than $i^2 = -1$i2=−1 and comparing components.

There is a second, philosophical, payoff. Because $\mathbb{C}$C is algebraically closed but $\mathbb{R}$R is not, the complex numbers are in a precise sense the "completion" of the reals for the purposes of algebra — and yet, geometrically, $\mathbb{C}$C is just the plane $\mathbb{R}^2$R2 with a multiplication bolted on. Holding those two pictures (algebraic closure and planar geometry) together is what makes the rest of this strand so powerful.

---

Additional A* practice (with worked answers)

P1. Simplify $i^{15} + i^{16} + i^{17} + i^{18}$i15+i16+i17+i18. Answer: The four consecutive powers are $i^3, i^0, i^1, i^2 = -i, 1, i, -1$i3,i0,i1,i2=−i,1,i,−1; sum $= 0$=0. (Any four consecutive powers of $i$i sum to $0$0.)

P2. Find real $p, q$p,q with $(2p + 1) + (q - 3)i = 7 - i$(2p+1)+(q−3)i=7−i. Answer: $2p + 1 = 7 \Rightarrow p = 3$2p+1=7⇒p=3; $q - 3 = -1 \Rightarrow q = 2$q−3=−1⇒q=2.

P3. Solve $z^2 - 2z + 10 = 0$z2−2z+10=0. Answer: $\Delta = 4 - 40 = -36$Δ=4−40=−36; $z = \dfrac{2 \pm 6i}{2} = 1 \pm 3i$z=22±6i​=1±3i. (Check: product $= 1 + 9 = 10$=1+9=10 ✓.)

P4. Find all real $x, y$x,y such that $(x + yi)^2 = -5 + 12i$(x+yi)2=−5+12i. Answer: $x^2 - y^2 = -5$x2−y2=−5 and $2xy = 12 \Rightarrow xy = 6$2xy=12⇒xy=6. Then $y = 6/x$y=6/x gives $x^2 - 36/x^2 = -5$x2−36/x2=−5, i.e. $x^4 + 5x^2 - 36 = 0$x4+5x2−36=0, so $x^2 = 4$x2=4 (reject $-9$−9), $x = \pm 2$x=±2, $y = \pm 3$y=±3. Square roots: $\pm(2 + 3i)$±(2+3i).

P5. Given $z^2 + 4z + 7 = 0$z2+4z+7=0, express the roots in the form $a + bi$a+bi and verify their product equals the constant term. Answer: $\Delta = 16 - 28 = -12$Δ=16−28=−12; $z = \dfrac{-4 \pm 2\sqrt{3}\,i}{2} = -2 \pm \sqrt{3}\,i$z=2−4±23​i​=−2±3​i. Product $= (-2)^2 + (\sqrt 3)^2 = 4 + 3 = 7$=(−2)2+(3​)2=4+3=7 ✓.

P6. Simplify $\dfrac{i^{25} + i^{-25}}{i^{50}}$i50i25+i−25​. Answer: $i^{25} = i^{1} = i$i25=i1=i; $i^{-25} = i^{-25 + 28} = i^{3} = -i$i−25=i−25+28=i3=−i; $i^{50} = i^{2} = -1$i50=i2=−1. So the expression is $\dfrac{i + (-i)}{-1} = \dfrac{0}{-1} = 0.$−1i+(−i)​=−10​=0.

P7. The equation $z^2 + kz + 4 = 0$z2+kz+4=0 (with $k$k real) has complex (non-real) roots. Find the set of values of $k$k. Answer: Non-real roots require $\Delta < 0$Δ<0: $k^2 - 16 < 0 \Rightarrow -4 < k < 4$k2−16<0⇒−4<k<4.

---

Board-alignment footer

Aligned to AQA A-Level Further Mathematics 7367, compulsory pure (Papers 1 & 2): definition of $i$i, the form $a + bi$a+bi, powers of $i$i, and real-coefficient quadratics with complex roots. The content is equivalent to the corresponding sections of Edexcel (9FM0) Core Pure and OCR (H245) — the algebra of $i$i is identical across boards.

---

Visual summary

$\boxed{\,i^2 = -1 \quad\Rightarrow\quad z = a + bi,\; a,b \in \mathbb{R};\quad i^n = i^{\,n \bmod 4};\quad a + bi = c + di \iff a = c,\ b = d\,}$i2=−1⇒z=a+bi,a,b∈R;in=inmod4;a+bi=c+di⟺a=c, b=d​

Idea	Key result
Defining property	$i^2 = -1$i2=−1
Standard form	$z = a + bi$z=a+bi, $\operatorname{Re}(z) = a$Re(z)=a, $\operatorname{Im}(z) = b$Im(z)=b
Powers of $i$i	period 4: $i, -1, -i, 1, \ldots$i,−1,−i,1,…
Surd of a negative	$\sqrt{-k} = \sqrt{k}\,i\ (k>0)$−k​=k​i (k>0)
Equality	compare real and imaginary parts
Real quadratic, $\Delta < 0$Δ<0	complex conjugate pair of roots

Recap. A single rule, $i^2 = -1$i2=−1, creates $\mathbb{C}$C; every complex number is $a + bi$a+bi; powers of $i$i repeat every four; and "equate real and imaginary parts" is the workhorse you will use in almost every problem that follows.