Complex Numbers and Quadratic Equations

The chapter has a clean, cumulative structure that this companion follows section by section. It opens by defining the complex number system itself — what $i$ is, what a complex number is, and what it means for two complex numbers to be equal. It then builds the algebra: addition, subtraction, multiplication and division, the field laws these operations obey, the periodic behaviour of powers of $i$, and the careful rule for taking square roots of negative real numbers (where a classic algebra trap lives). Next come the modulus and conjugate, the two tools that measure a complex number's size and let you divide by it. The chapter then turns geometric with the Argand plane and the polar (trigonometric) representation of a complex number as a length $r$ and an angle $\theta$. Finally, it closes the loop opened in the introduction: quadratic equations with real coefficients and negative discriminant, now fully solvable using $i$. Work through the sections in this order — each one is the toolkit for the next.

Defining $i$ and the complex number system

We begin by naming the missing square root. Let $i$ denote a number satisfying

where: $i$ is called the imaginary unit; it is defined by this property, not derived from anything more basic. A complex number is then any expression of the form $z = a + ib$, where $a$ and $b$ are real numbers. The real number $a$ is called the real part of $z$, written $\text{Re}\, z = a$, and the real number $b$ is called the imaginary part, written $\text{Im}\, z = b$. Note carefully that $\text{Im}\, z$ is the real number $b$, not $ib$. For example, if $z = 2 + 5i$, then $\text{Re}\, z = 2$ and $\text{Im}\, z = 5$. A complex number with $b = 0$ is just an ordinary real number, so the real numbers sit inside the complex numbers as the special case of zero imaginary part; a complex number with $a = 0$ (and $b \neq 0$) is called purely imaginary. Two complex numbers are declared equal only when they match feature for feature:

where: this is not a theorem to prove but the working definition of equality — a single complex equation secretly carries two independent real conditions, one from matching real parts and one from matching imaginary parts. This is exactly the trick used to solve for two unknowns hidden inside one complex equation, as in NCERT's own Example 1: from $4x + i(3x-y) = 3 + i(-6)$, equating real parts gives $4x = 3$ and equating imaginary parts gives $3x - y = -6$, a pair of ordinary simultaneous equations solving to $x = 3/4$ and $y = 33/4$.

The algebra of complex numbers

Addition, subtraction and multiplication of complex numbers are defined so that ordinary algebra "just works" once you remember $i^2 = -1$. For $z_1 = a+ib$ and $z_2 = c+id$:

where: addition simply adds real parts and imaginary parts separately; multiplication is what you get by expanding $(a+ib)(c+id) = ac + iad + ibc + i^2bd$ and replacing $i^2$ by $-1$, which turns $bd$ into $-bd$ and collects the imaginary terms $ad+bc$. Subtraction is addition of the additive inverse: $z_1 - z_2 = z_1 + (-z_2)$, where $-z_2 = -c - id$. These operations obey the same laws you already trust for real numbers: the closure law (the sum and product of two complex numbers are complex numbers), the commutative law ($z_1+z_2=z_2+z_1$ and $z_1z_2=z_2z_1$), the associative law for both operations, and the distributive law $z_1(z_2+z_3) = z_1z_2 + z_1z_3$. There is an additive identity $0 = 0+i0$ with $z+0=z$, and every $z$ has an additive inverse $-z$ with $z+(-z)=0$. There is a multiplicative identity $1 = 1+i0$ with $z\cdot 1 = z$, and every non-zero $z$ has a multiplicative inverse $z^{-1}$ with $z\cdot z^{-1}=1$ — its formula is derived below, using the modulus and conjugate. Division is then simply $\dfrac{z_1}{z_2} = z_1 \cdot z_2^{-1}$ for $z_2 \neq 0$. Because all these familiar laws survive intact, algebraic identities that hold for real numbers — such as $(z_1+z_2)^2 = z_1^2 + 2z_1z_2 + z_2^2$, $(z_1-z_2)^2 = z_1^2 - 2z_1z_2 + z_2^2$, the cube expansions $(z_1\pm z_2)^3 = z_1^3 \pm 3z_1^2z_2 + 3z_1z_2^2 \pm z_2^3$, and the factorisation $z_1^2 - z_2^2 = (z_1+z_2)(z_1-z_2)$ — continue to hold for complex numbers, proved by exactly the same distributive-law algebra as in the real case.

Powers of $i$ and square roots of negative numbers

Because $i^2 = -1$, every higher power of $i$ reduces to one of just four values. Compute $i^3 = i^2 \cdot i = -i$ and $i^4 = (i^2)^2 = (-1)^2 = 1$; then $i^5 = i^4 \cdot i = i$ repeats the cycle. In general, for any integer $k$,

where: to evaluate $i^n$ for any integer $n$, divide $n$ by 4 and read off the remainder — the answer only ever depends on $n \bmod 4$. Negative powers reduce the same way, using $i^{-1} = 1/i = -i$ (multiply numerator and denominator by $i$: $1/i = i/i^2 = i/(-1) = -i$); for instance $i^{-4} = 1/i^4 = 1/1 = 1$. For a positive real number $a$, the symbol $\sqrt{-a}$ is defined to mean $i\sqrt{a}$, so that $(i\sqrt a)^2 = i^2 a = -a$ as required. This is where the chapter issues its sharpest warning: the familiar real-number rule $\sqrt{a}\times\sqrt{b} = \sqrt{ab}$ does not survive when both $a$ and $b$ are negative. If it did, taking $a=b=-1$ would give $\sqrt{-1}\times\sqrt{-1} = \sqrt{(-1)(-1)} = \sqrt{1} = 1$; but the left side is actually $i \times i = i^2 = -1$, a direct contradiction. The rule $\sqrt a \sqrt b = \sqrt{ab}$ remains valid whenever at least one of $a,b$ is non-negative; it must be abandoned, and the two square roots evaluated separately as $i\sqrt{|a|}$ and $i\sqrt{|b|}$ first, whenever both are negative.

Modulus and conjugate

For $z = a+ib$, define the modulus $|z| = \sqrt{a^2+b^2}$, a non-negative real number, and the conjugate $\bar z = a - ib$, obtained by flipping the sign of the imaginary part. For example, $|3+i| = \sqrt{3^2+1^2} = \sqrt{10}$ and $\overline{3+i} = 3-i$. These two quantities are linked by the identity proved in the first derivation below:

where: this single identity is the engine behind division of complex numbers — it converts "dividing by $z$" into "multiplying by $\bar z$ and dividing by the real number $|z|^2$," which is exactly how you rationalise a complex denominator. The modulus and conjugate obey clean multiplicative rules for any two complex numbers $z_1, z_2$: $\overline{z_1 z_2} = \bar z_1 \bar z_2$, $\overline{z_1 \pm z_2} = \bar z_1 \pm \bar z_2$, $|z_1 z_2| = |z_1||z_2|$ (proved in the third derivation below), and $\left|\dfrac{z_1}{z_2}\right| = \dfrac{|z_1|}{|z_2|}$ for $z_2 \neq 0$.

The Argand plane, geometry of the conjugate, and polar form

Plotting $z = x+iy$ as the point $P(x,y)$ in a plane with a horizontal real axis and vertical imaginary axis turns algebra into geometry; this plane is called the Argand plane or complex plane. The modulus $|z| = \sqrt{x^2+y^2}$ is then simply the distance $OP$ from the origin — the ordinary distance formula in disguise. The conjugate $\bar z = x - iy$ plots as the point $Q(x,-y)$, the mirror image of $P$ across the real axis; geometrically, conjugation is reflection. Every non-zero point $P(x,y)$ can also be located by a length and an angle instead of two coordinates. Write $x = r\cos\theta$ and $y = r\sin\theta$, where $r = OP$; then

where: this is the polar form (or trigonometric form) of $z$; $r = |z| = \sqrt{x^2+y^2}$ is the modulus exactly as before, and $\theta$, called the argument (or amplitude) of $z$, is the angle $OP$ makes with the positive real axis, measured anticlockwise. Any $\theta$ satisfying $\cos\theta = x/r$ and $\sin\theta = y/r$ works, so the argument is only determined up to adding multiples of $2\pi$; the value chosen from the interval $(-\pi, \pi]$ is called the principal argument. For example, $z = 1+i$ has $r = \sqrt2$ and lies in the first quadrant with $\cos\theta = \sin\theta = 1/\sqrt2$, so its principal argument is $\theta = \pi/4$, giving the polar form $z = \sqrt2\left(\cos\dfrac{\pi}{4} + i\sin\dfrac{\pi}{4}\right)$.

Complex roots of a quadratic equation

We close the loop opened at the start of the chapter. For a quadratic equation $ax^2+bx+c=0$ with real coefficients $a,b,c$ ($a\neq 0$) and discriminant $D = b^2-4ac$, the familiar quadratic formula gives real roots when $D \geq 0$; when $D < 0$, $\sqrt D$ does not exist among the reals, but it exists as a complex number, and the roots are

where: $4ac-b^2 = -D$ is now a positive real number, so $\sqrt{4ac-b^2}$ is an ordinary real square root, and the two roots are complex conjugates of each other, symmetric about the real number $-b/2a$. The full derivation, by completing the square, is given below; the key qualitative fact worth remembering is that a quadratic with real coefficients and negative discriminant never has just one complex root — complex roots of real-coefficient polynomials always arrive in conjugate pairs.

Board level - Finding the multiplicative inverse of a complex number

Find the multiplicative inverse of $z = 2 - 3i$, and verify the answer.

Step 1. Identify $a=2$, $b=-3$, so $\bar z = 2+3i$ and $|z|^2 = a^2+b^2 = 2^2+(-3)^2 = 4+9=13$.

Step 2. Apply the formula from the first derivation, $z^{-1} = \bar z/|z|^2$: $z^{-1} = \dfrac{2+3i}{13} = \dfrac{2}{13} + \dfrac{3}{13}i$.

Step 3. Verify by multiplying $z \cdot z^{-1}$ directly: $(2-3i)\left(\dfrac{2}{13}+\dfrac{3}{13}i\right) = \dfrac{1}{13}\left[(2-3i)(2+3i)\right] = \dfrac{1}{13}\left(4 - 9i^2\right) = \dfrac{1}{13}(4+9) = \dfrac{13}{13} = 1$, confirming the identity is correct. Answer: $z^{-1} = \dfrac{2}{13} + \dfrac{3}{13}i$.

Competitive level (JEE / NEET) - Solving a quadratic with complex roots and checking Vieta's formulas

Solve $2x^2 - 4x + 5 = 0$ for $x$, and verify that the sum and product of the roots match $-b/a$ and $c/a$.

Step 1. Compute the discriminant: $D = b^2-4ac = (-4)^2 - 4(2)(5) = 16-40 = -24$. Since $D<0$, the roots are complex, and $4ac-b^2 = 24$.

Step 2. Substitute into formula (7): $x = \dfrac{-b \pm i\sqrt{4ac-b^2}}{2a} = \dfrac{4 \pm i\sqrt{24}}{4} = \dfrac{4 \pm 2i\sqrt 6}{4} = 1 \pm \dfrac{i\sqrt6}{2}$.

Step 3. Check Vieta's formulas. Sum of roots: $\left(1+\dfrac{i\sqrt6}{2}\right)+\left(1-\dfrac{i\sqrt6}{2}\right) = 2$, and $-b/a = -(-4)/2 = 2$; they match. Product of roots, using $z\bar z=|z|^2$ from the first derivation with $z = 1+\dfrac{i\sqrt6}{2}$: $|z|^2 = 1^2+\left(\dfrac{\sqrt6}{2}\right)^2 = 1+\dfrac{6}{4} = \dfrac{5}{2}$, and $c/a = 5/2$; they match too. Answer: $x = 1 \pm \dfrac{i\sqrt6}{2}$, and both Vieta checks confirm the roots. Timing note: under ninety seconds. Compute $D$ first — its sign alone tells you which formula branch to use, so never substitute into the real-root formula before checking the sign of $D$. The Vieta check ($\alpha+\beta=-b/a$, $\alpha\beta=c/a$) is the fastest self-check available and catches almost every arithmetic slip in a JEE-speed solution; skipping it is the single most common source of silently wrong answers on this topic.

Competitive level (JEE / NEET) - Locus of a point in the Argand plane from a modulus condition

If $z=x+iy$ is a complex number satisfying $|z-1| = |z+i|$, find the equation of the locus traced by the point $(x,y)$, and identify the curve.

Step 1. Write $z-1 = (x-1)+iy$ and $z+i = x+i(y+1)$, splitting the condition into real and imaginary parts using the definitions of modulus, so that $|z-1| = \sqrt{(x-1)^2+y^2}$ and $|z+i| = \sqrt{x^2+(y+1)^2}$.

Step 2. Since both sides are non-negative, square the given equation directly: $(x-1)^2+y^2 = x^2+(y+1)^2$.

Step 3. Expand both sides: left side is $x^2-2x+1+y^2$; right side is $x^2+y^2+2y+1$. Cancel the common terms $x^2+y^2+1$ from both sides, leaving $-2x = 2y$, which simplifies to $y = -x$, or $x+y=0$.

Step 4. Interpret geometrically: the condition $|z-1|=|z+i|$ says the point $z$ is equidistant from the two fixed points $1+i0$ and $0-i$ in the Argand plane, so by the standard geometric fact that such points form the perpendicular bisector of the segment joining them, the locus is a straight line — confirming the algebra. Answer: the locus is the straight line $x+y=0$, the perpendicular bisector of the segment joining $(1,0)$ and $(0,-1)$. Timing note: about two minutes. Recognising "$|z-a|=|z-b|$ is the perpendicular bisector of $a$ and $b$" instantly, without expanding anything, is the fast JEE route; the full algebraic expansion above is the safety-net method when the two fixed points are not obviously recognised, or when the exam wants the working shown.

Almost every recurring error in this chapter comes from applying a real-number habit to $i$ without checking it still holds. Watch for these specifically:

• $\sqrt a \sqrt b \neq \sqrt{ab}$ when both $a,b$ are negative. Always convert each negative root to $i\sqrt{|a|}$ first, then multiply; never combine two negative radicands under one root sign.
• Confusing $\text{Im}\,z$ with $ib$. The imaginary part of $z=a+ib$ is the real number $b$, not $ib$; writing $\text{Im}\,z = ib$ is a very common slip that then poisons every equality-of-parts calculation.
• Forgetting that $z\bar z$ is real, not complex. Students sometimes leave a stray $i$ in a modulus-squared calculation; the whole point of Step 2 of the first derivation is that the imaginary cross-terms cancel exactly.
• Applying the real quadratic formula's "no solution" verdict. When $D<0$, the equation is not unsolvable — it has two genuine complex roots; "no real roots" and "no roots" are different statements, and only the first is ever true here.
• Mixing up modulus and argument conventions. The principal argument is conventionally taken in $(-\pi,\pi]$, not $[0,2\pi)$; using the wrong convention shifts negative-imaginary-part answers by $2\pi$ and is a frequent JEE penalty.
• Rationalising by the wrong conjugate. Dividing $z_1/z_2$ requires multiplying numerator and denominator by $\bar z_2$ (the conjugate of the denominator), not $\bar z_1$; the goal is to make the denominator real using $z_2\bar z_2=|z_2|^2$.
• Treating powers of $i$ as needing full multiplication every time. Reduce the exponent modulo 4 first; multiplying out $i^{47}$ term by term instead of noting $47 = 4(11)+3 \Rightarrow i^{47}=i^3=-i$ wastes time and invites arithmetic errors.

• Define a complex number, identify its real and imaginary parts, and correctly apply the equality condition to split one complex equation into two real equations.
• Add, subtract, multiply and divide complex numbers, and state the closure, commutative, associative and distributive laws they obey.
• Reduce any power of $i$, positive or negative, to one of $\{1, i, -1, -i\}$ using the exponent-modulo-4 rule.
• Correctly take the square root of a negative real number and avoid the $\sqrt a\sqrt b=\sqrt{ab}$ trap when both quantities are negative.
• Compute the modulus and conjugate of any complex number, and prove and apply the identity $z\bar z=|z|^2$.
• Derive and apply the multiplicative inverse formula $z^{-1}=\bar z/|z|^2$, and use it to divide complex numbers by rationalising the denominator.
• Plot a complex number in the Argand plane, and convert freely between Cartesian form $x+iy$ and polar form $r(\cos\theta+i\sin\theta)$, including finding the principal argument in each quadrant.
• Derive the quadratic formula by completing the square and extend it correctly to the case $D<0$, producing a conjugate pair of complex roots.
• Verify Vieta's formulas (sum and product of roots) for a quadratic with complex roots, and recall that real-coefficient quadratics never have a lone complex root.
• Solve modulus-based locus problems in the Argand plane, recognising conditions like $|z-a|=|z-b|$ geometrically as a perpendicular bisector.

Ask the tutor: Want to work through a few more modulus-and-argument or complex-quadratic problems together to build exam speed?