Complex Numbers and Quadratic Equations

Use the familiar rule $\sqrt{p}\sqrt{q}=\sqrt{pq}$ on a single innocent product.

$\sqrt{-1}\sqrt{-1}=\sqrt{(-1)(-1)}=\sqrt{1}=1$

But $\sqrt{-1}$ is $i$, and $i\times i=i^2=-1$.

The rule has just proved that $1=-1$.

Nothing is wrong with $i$. What is wrong is the phrase "the square root of $-1$". Every non-zero number has two square roots, and for positives we agree to mean the positive one. For negatives there is no such agreement available, because neither $i$ nor $-i$ is larger.

That is why the syllabus does not define $i$ as $\sqrt{-1}$. It defines a complex number as an ordered pair $(a,b)$ with a stated multiplication rule, and then $i^2=-1$ comes out as a consequence rather than a declaration.

Once the plane exists, the payoff arrives immediately.

Take the first. $(1+i{)}^2=2i$, so $(1+i{)}^{2026}=(2i{)}^{1013}=2^{1013}{i}^{1013}$, and $1013$ leaves remainder $1$ on division by $4$, giving $2^{1013}i$.

Never compute what you can describe. A complex number is described by modulus and argument, because that turns multiplication into rotation. A quadratic's roots are described by their sum and product, because the coefficients hand you both for free.

1. Why Complex Numbers Exist

$x^2+1=0$ has no real solution, since a real square is never negative. Extending the number system to fix that is the same move that produced negatives and then fractions.

Definition. A complex number is an ordered pair of reals $(a,b)$, added componentwise, multiplied by the rule below.

$(a,b)(c,d)=(ac-bd,ad+bc)$

Set $i=(0,1)$ and apply the rule to itself: $(0,1)(0,1)=(0\cdot 0-1\cdot 1,0\cdot 1+1\cdot 0)=(-1,0)$.

$i^2=-1$

Every complex number is then written $a+ib$, with real part $a$ and imaginary part $b$.

Trap. The imaginary part of $3-5i$ is the real number $-5$, not $-5i$.

The working rule. Convert every negative under a root to $i$ form first, then multiply.

$\sqrt{-4}\sqrt{-9}=(2i)(3i)=6i^2=-6$

2. Algebra of Complex Numbers

Addition and subtraction act componentwise. Multiplication is ordinary expansion with $i^2$ replaced by $-1$.

Conjugate. $\overline{z}=a-ib$ for $z=a+ib$: flip the sign of the imaginary part. Geometrically, a reflection in the real axis.

$z\overline{z}=(a+ib)(a-ib)=a^2+b^2=|z{|}^2$

That identity is the workhorse of the chapter, because it turns a complex product into a real number. It is exactly why division works.

Illustration 1

Write $\frac{3+4i}{1-2i}$ in the form $a+ib$.

Multiply top and bottom by the conjugate of the denominator, which makes the denominator real.

$\frac{3+4i}{1-2i}\times \frac{1+2i}{1+2i}=\frac{3+6i+4i+8i^2}{1+4}=\frac{-5+10i}{5}=-1+2i$

The denominator became $1^2+2^2=5$ by the $z\overline{z}=|z{|}^2$ identity, with no expansion needed there at all.

Check with moduli, which is faster than re-expanding: $|3+4i|=5$ and $|1-2i|=\sqrt{5}$, so the quotient must have modulus $5/\sqrt{5}=\sqrt{5}$. And $|-1+2i|=\sqrt{5}$.

Equality, and the absence of order

Two complex numbers are equal exactly when real parts match and imaginary parts match. One equation in $\mathbb{C}$ is therefore two equations in $\mathbb{R}$, which is how most "find $x$ and $y$" questions are solved.

Trap. There is no consistent order on $\mathbb{C}$. Writing $z_1<z_2$ is meaningless, and $i>0$ leads to contradictions either way. Only moduli may be compared, because those are real.

3. The Argand Plane: Modulus and Argument

Plotting $a+ib$ at $(a,b)$ gives the Argand plane, and the chapter becomes geometry.

$|z|=\sqrt{a^2+b^2}$

Modulus is distance from the origin, so $|z_1-z_2|$ is the distance between two points. Every locus question uses that one reading.

Argument is the angle to the positive real axis, measured anticlockwise. The principal argument lies in $(-\pi ,\pi ]$.

Trap. The argument is not $\arctan (b/a)$. The arctangent function returns values in $(-\pi /2,\pi /2)$ only, so it cannot distinguish the second quadrant from the fourth, nor the third from the first.

$z=r(\cos \theta +i\sin \theta ),r=|z|$

Illustration 2

Find the principal arguments of $z_1=-1+\sqrt{3}i$ and $z_2=1-\sqrt{3}i$.

Both give $b/a=-\sqrt{3}$, so a calculator returns $-60^{\circ }$ for both. At most one of those can be right.

$z_1$ lies in the second quadrant, where the real part is negative and the imaginary part positive. Reference angle $\arctan (\sqrt{3})=60^{\circ }$, so the argument is $180^{\circ }-60^{\circ }=120^{\circ }$, that is $2\pi /3$.

$z_2$ lies in the fourth quadrant. Same reference angle, so the argument is $-60^{\circ }$, that is $-\pi /3$.

$\arg z_1=\frac{2\pi }{3},\arg z_2=-\frac{\pi }{3}$

The two differ by $\pi$, as they must, because $z_2=-z_1$. Negating a complex number turns it through a straight angle, and a ratio $b/a$ cannot see that turn at all.

4. Multiplication Is Rotation

This is the single most useful fact in the chapter.

$z_1{z}_2=r_1{r}_2\left[\cos ({\theta }_1+{\theta }_2)+i\sin ({\theta }_1+{\theta }_2)\right]$

Read it as an instruction: multiply the moduli, add the arguments.

$|z_1{z}_2|=|z_1||z_2|,\arg (z_1{z}_2)=\arg z_1+\arg z_2(\text{mod }2\pi )$

Multiplying by $i$, which has modulus $1$ and argument $\pi /2$, is therefore a quarter turn about the origin with no change of distance.

De Moivre's theorem is the same rule applied $n$ times.

$z^n=r^n(\cos n\theta +i\sin n\theta )$

To rotate about a point other than the origin: subtract it, multiply, add it back.

$z^{\prime }=z_0+(z-z_0)(\cos \alpha +i\sin \alpha )$

Trap. Arguments add only up to multiples of $2\pi$. If the sum leaves $(-\pi ,\pi ]$, adjust by $2\pi$ before quoting a principal argument. JEE has tested exactly that gap.

Illustration 3

$z_1=1$ and $z_2=3$ are two vertices of an equilateral triangle. Find a third vertex.

Coordinates would need the perpendicular bisector, the height $\sqrt{3}$, and care with signs. Rotation needs one line.

Rotate $z_2$ about $z_1$ through $60^{\circ }$.

$z_3=1+(3-1)\left(\cos 60^{\circ }+i\sin 60^{\circ }\right)=1+2\left(\frac{1}{2}+\frac{\sqrt{3}}{2}i\right)=2+\sqrt{3}i$

Check both new sides. $|z_3-z_1|=|1+\sqrt{3}i|=2$ and $|z_3-z_2|=|-1+\sqrt{3}i|=2$, matching $|z_2-z_1|=2$.

Rotating through $-60^{\circ }$ instead gives $2-\sqrt{3}i$, the reflection below the axis. Both are correct, and a question asking for "the" third vertex has usually restricted the half-plane.

Illustration 4

Find the smallest positive integer $n$ for which ${\left(\frac{1+i}{1-i}\right)}^n=1$.

Simplify the base before raising anything to a power.

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

So the question is when $i^n=1$, and the powers of $i$ cycle with period $4$: $i,-1,-i,1$.

$n=4$

The description says the same thing without arithmetic. The numerator has argument $\pi /4$ and the denominator $-\pi /4$, so the quotient has argument $\pi /2$ and modulus $1$. A quarter turn returns to the start after four applications.

Roots of unity

De Moivre's theorem run backwards solves $z^n=1$. Modulus $1$ and argument a multiple of $2\pi /n$ is the only possibility.

$z_k=\cos \frac{2k\pi }{n}+i\sin \frac{2k\pi }{n},k=0,1,\dots ,n-1$

So the $n$-th roots of unity are $n$ equally spaced points on the unit circle, one of them always at $1$.

For $n=3$ they are written $1$, $\omega$ and ${\omega }^2$.

$\omega =-\frac{1}{2}+\frac{\sqrt{3}}{2}i,{\omega }^3=1,1+\omega +{\omega }^2=0$

The last identity is the one that does the work, and it is geometry rather than algebra: three unit vectors at $120^{\circ }$ to one another cancel out. The same argument shows the $n$-th roots of unity sum to zero for every $n\ge 2$.

Illustration 5

Evaluate ${\left(1+\omega -{\omega }^2\right)}^7$, where $\omega$ is a non-real cube root of unity.

Expanding is hopeless. Use $1+\omega +{\omega }^2=0$ to collapse the bracket first.

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

${\left(-2{\omega }^2\right)}^7=-128{\omega }^{14}$

Now reduce the exponent using ${\omega }^3=1$: $14=3(4)+2$, so ${\omega }^{14}={\omega }^2$.

${\left(1+\omega -{\omega }^2\right)}^7=-128{\omega }^2$

Every step replaced a power by a smaller one. That is the whole technique with $\omega$: reduce the exponent modulo $3$, and use the sum identity to remove any $1$ that appears alongside $\omega$ terms.

5. Modulus Inequalities and Loci

Since $|z_1-z_2|$ is a distance, a modulus condition is a distance condition and the locus is whatever curve satisfies it.

$|z_1+z_2|\le |z_1|+|z_2|,||z_1|-|z_2||\le |z_1+z_2|$

The triangle inequality carries over unchanged, because the moduli really are side lengths. Equality holds exactly when the arguments agree, that is when the two point the same way. Together the pair brackets a modulus whenever a question asks for a greatest or least value.

Illustration 6

Find the locus of $z$ with $|z-1|=2|z+1|$.

The instinct is that equal-looking distance conditions give lines. Square both sides and see.

$(x-1{)}^2+y^2=4\left[(x+1{)}^2+y^2\right]$

$x^2-2x+1+y^2=4x^2+8x+4+4y^2\Rightarrow 3x^2+3y^2+10x+3=0$

$x^2+y^2+\frac{10}{3}x+1=0\Rightarrow {\left(x+\frac{5}{3}\right)}^2+y^2=\frac{25}{9}-1=\frac{16}{9}$

A circle of centre $(-5/3,0)$ and radius $4/3$, called an Apollonius circle.

The $x^2$ and $y^2$ terms are what decide it. When $k=1$ they cancel between the two sides and a line survives; for any other $k$ they cannot cancel, so the locus is forced to be a circle.

Sanity check the two points on the real axis: $x=-1/3$ gives $|-1/3-1|=4/3$ and $2|-1/3+1|=4/3$, and $x=-3$ gives $4$ and $2\times 2=4$. Both lie at distance $4/3$ from $-5/3$.

Illustration 7

If $\left|z+\frac{1}{z}\right|=2$, find the greatest and least values of $|z|$.

Nothing here identifies $z$, and nothing needs to. Write $r=|z|$ and squeeze it between the two inequalities.

$r=\left|\left(z+\frac{1}{z}\right)-\frac{1}{z}\right|\le \left|z+\frac{1}{z}\right|+\frac{1}{r}=2+\frac{1}{r}$

$r^2-2r-1\le 0\Rightarrow r\le 1+\sqrt{2}$

Running the reverse inequality from the other side gives the lower bound.

$\frac{1}{r}-r\le \left|z+\frac{1}{z}\right|=2\Rightarrow r^2+2r-1\ge 0\Rightarrow r\ge \sqrt{2}-1$

$|z{|}_{\max }=\sqrt{2}+1,|z{|}_{\min }=\sqrt{2}-1$

The two bounds are reciprocals, which they must be: replacing $z$ by $1/z$ leaves the condition unchanged, so the set of allowed moduli is closed under $r\mapsto 1/r$.

6. Quadratic Equations: Nature of Roots

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a},D=b^2-4ac$

Everything about the roots, without computing them, is read off the discriminant.

Trap. Two rows of that table carry conditions, and dropping them is the commonest error here. Complex roots pair up only when the coefficients are real: $x^2-(2+i)x+2i=0$ has roots $2$ and $i$. Irrational roots pair up only when the coefficients are rational.

Solving in the complex number system

With real coefficients and $D<0$ the formula still applies; the square root simply lands off the real line.

$x^2+4x+13=0\Rightarrow D=16-52=-36\Rightarrow \sqrt{D}=6i\Rightarrow x=-2\pm 3i$

Nothing structural changed. Sum $=-4=-b/a$; product $=4+9=13=c/a$. That stability is the point of extending the number system: every quadratic now has exactly two roots, with no exceptions to remember.

Illustration 8

For which real $m$ does $x^2-2(m+1)x+m^2+3=0$ have real roots?

Real roots means $D\ge 0$, and nothing else is required.

$D=4(m+1{)}^2-4(m^2+3)=4\left[m^2+2m+1-m^2-3\right]=8(m-1)$

$8(m-1)\ge 0\Rightarrow m\ge 1$

The $m^2$ terms cancelled, which is why the answer is a half-line rather than an interval. At $m=1$ the discriminant vanishes and the roots coincide at $x=2$, so if the question had demanded distinct real roots the answer would be $m>1$.

7. Relation Between Roots and Coefficients

$\alpha +\beta =-\frac{b}{a},\alpha \beta =\frac{c}{a}$

This is the heart of the organising principle. Most questions ask for a symmetric combination of the roots, and every symmetric combination rewrites in terms of the sum and the product.

Illustration 9

If $\alpha ,\beta$ are the roots of $x^2-5x+3=0$, find ${\alpha }^4+{\beta }^4$.

The roots are $(5\pm \sqrt{13})/2$, so the direct route means raising a surd to the fourth power twice. Build up from the symmetric functions instead.

$\alpha +\beta =5,\alpha \beta =3$

${\alpha }^2+{\beta }^2=5^2-2(3)=19$

Now repeat the same identity one level up, treating ${\alpha }^2$ and ${\beta }^2$ as the two quantities.

${\alpha }^4+{\beta }^4={\left({\alpha }^2+{\beta }^2\right)}^2-2(\alpha \beta {)}^2=19^2-2(9)=361-18=343$

Exactly $343$, which is $7^3$. The decimal route gives $4.3028^4+0.6972^4\approx 342.76+0.236$, and rounding at any step would never have revealed that the answer is a whole number at all.

8. Forming Quadratics and Common Roots

$x^2-(\text{sum})x+(\text{product})=0$

Questions that transform the roots are answered by computing the new sum and product from the old ones, never by finding the roots.

Illustration 10

$\alpha ,\beta$ are the roots of $x^2-6x+4=0$. Form the equation whose roots are $\alpha /\beta$ and $\beta /\alpha$.

Both required roots are symmetric in $\alpha$ and $\beta$ as a pair, so both the new sum and the new product come straight from the old ones.

$\text{new sum}=\frac{\alpha }{\beta }+\frac{\beta }{\alpha }=\frac{{\alpha }^2+{\beta }^2}{\alpha \beta }=\frac{36-8}{4}=7$

$\text{new product}=\frac{\alpha }{\beta }\cdot \frac{\beta }{\alpha }=1$

$x^2-7x+1=0$

The product being $1$ is a structural check rather than a coincidence: the two new roots are reciprocals of each other, so any equation with them must have equal first and last coefficients.

Common roots

Both roots common means the equations are proportional.

$\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$

Exactly one root common gives the eliminant below.

$(c_1{a}_2-c_2{a}_1{)}^2=(a_1{b}_2-a_2{b}_1)(b_1{c}_2-b_2{c}_1)$

In practice, with numbers rather than letters, subtract the equations. The quadratic terms cancel when the leading coefficients match, leaving a linear equation that hands you the root.

Illustration 11

Find the common root of $x^2-5x+6=0$ and $x^2-9x+14=0$.

The eliminant would work, but subtraction is faster and needs no memory.

$\left(x^2-5x+6\right)-\left(x^2-9x+14\right)=4x-8=0\Rightarrow x=2$

Verify in both: $4-10+6=0$ and $4-18+14=0$. The common root is $2$.

Now note what subtraction actually proved. Any shared root must satisfy the difference, so it can only be $2$; the check is what confirms $2$ genuinely is a root rather than merely the sole candidate. Skipping the verification is how a question with no common root catches people out.

Summary

The rule $\sqrt{p}\sqrt{q}=\sqrt{pq}$ fails for negatives, which is why $i$ is defined by an ordered-pair multiplication rule and $i^2=-1$ is derived. Convert to $i$ form before multiplying.

$z\overline{z}=|z{|}^2$ turns a complex product real, which is why division multiplies by the conjugate.

There is no order on $\mathbb{C}$; only moduli may be compared. One complex equation is two real equations.

The argument is not $\arctan (b/a)$: plot the point, take the reference angle, then adjust by quadrant.

Multiplication multiplies moduli and adds arguments, so multiplying by $i$ is a quarter turn. De Moivre's theorem is that rule applied $n$ times, and it makes $(1+i{)}^{2026}=2^{1013}i$ a one-line answer.

Rotation about a point: subtract it, multiply by $\cos \alpha +i\sin \alpha$, add it back.

The $n$-th roots of unity are $n$ equally spaced points on the unit circle and always sum to zero. For cube roots, ${\omega }^3=1$ and $1+\omega +{\omega }^2=0$ reduce any expression in $\omega$ to a first-degree one.

$|z_1-z_2|$ is a distance, so every locus question is geometry. Equal distances give the perpendicular bisector; any other ratio gives a circle, because the squared terms no longer cancel.

The triangle inequality and its reverse bracket a modulus, which is how greatest and least values are found without identifying $z$.

The discriminant settles the nature of the roots, but conjugate pairing needs real coefficients and surd pairing needs rational ones.

Sum is $-b/a$ and product is $c/a$. Every symmetric combination rewrites in those two, so ${\alpha }^4+{\beta }^4=343$ falls out of $19^2-2(9)$ while the roots stay unknown.

Form an equation as $x^2$ minus the sum times $x$ plus the product, computing the new sum and product from the old.

For a numerical common root, subtract the two equations to get a linear one, then verify the candidate in both.