Easy Practice — Set 1

Quick Recap — Complex Numbers (Basics)

• $i^2=-1$; powers of $i$ cycle with period 4: $i,-1,-i,1$.
• For $z=a+bi$: conjugate $\bar z=a-bi$, modulus $|z|=\sqrt{a^2+b^2}$, and $z\bar z=|z|^2$.
• $z+\bar z=2a$ (real), $z-\bar z=2bi$ (imaginary).
• $|z_1z_2|=|z_1||z_2|$ and $\arg(z_1z_2)=\arg z_1+\arg z_2$.

Beyond-NCERT JEE Formulae

Past the NCERT core (cube roots of unity and the basic sum/product of roots are assumed known), these are the high-yield tools for JEE Main geometry-flavoured and root-location questions.

1. Rotation theorem

To rotate the vector $\vec{PQ}$ about the point $P$ through an angle $\theta$ onto $\vec{PR}$, where $P,Q,R$ correspond to $z_1,z_2,z_3$: $$\dfrac{z_3-z_1}{z_2-z_1}=\dfrac{|z_3-z_1|}{|z_2-z_1|}\,e^{i\theta}.$$ Multiplying any complex number $z$ by $e^{i\theta}$ rotates its position vector about the origin by $\theta$ (anticlockwise when $\theta>0$). [JEE Tip] For a square or rhombus the two sides are equal, so the modulus ratio is $1$ and the factor is just $e^{i\theta}$; a right-angle turn uses $e^{i\pi/2}=i$, and a $60$ degree turn (equilateral triangle) uses $e^{i\pi/3}=\tfrac12+\tfrac{\sqrt3}{2}i$.

2. Triangle inequality (two-sided)

$$\big||z_1|-|z_2|\big|\le |z_1\pm z_2|\le |z_1|+|z_2|.$$ The upper bound is attained when $z_1,z_2$ point the same way (equal argument), the lower bound when they are oppositely directed. This is the standard engine for the greatest/least value of $|z|$ under a constraint.

3. Parallelogram law and the Re-expansion

$$|z_1+z_2|^2+|z_1-z_2|^2=2\big(|z_1|^2+|z_2|^2\big),$$ and, splitting a single modulus, $$|z_1\pm z_2|^2=|z_1|^2+|z_2|^2\pm 2\operatorname{Re}(z_1\bar z_2).$$ The first identity reads: the sum of the squares of the diagonals equals the sum of the squares of the four sides.

4. $n$th roots of unity

The solutions of $z^n=1$ are $1,\alpha,\alpha^2,\dots,\alpha^{n-1}$ with $\alpha=e^{2\pi i/n}$, and they satisfy $$1+\alpha+\alpha^2+\cdots+\alpha^{n-1}=0,\qquad 1\cdot\alpha\cdot\alpha^2\cdots\alpha^{n-1}=(-1)^{n-1}.$$ Geometrically they are the vertices of a regular $n$-gon inscribed in the unit circle. Since $\displaystyle\prod_{k=1}^{n-1}(x-\alpha^k)=1+x+x^2+\cdots+x^{n-1}$, putting $x=1$ gives the handy $\displaystyle\prod_{k=1}^{n-1}(1-\alpha^k)=n$.

5. Common roots of two quadratics

For $a_1x^2+b_1x+c_1=0$ and $a_2x^2+b_2x+c_2=0$:

• Exactly one common root $\Rightarrow (c_1a_2-c_2a_1)^2=(a_1b_2-a_2b_1)(b_1c_2-b_2c_1)$, and that root equals $\dfrac{c_1a_2-c_2a_1}{a_1b_2-a_2b_1}$.
• Both roots common $\Rightarrow \dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}=\dfrac{c_1}{c_2}$ (the two equations are proportional).

6. Location of the roots of $f(x)=ax^2+bx+c$ relative to a number $k$

Let $D=b^2-4ac$ and let the vertex sit at $x=-\dfrac{b}{2a}$.

• Both roots $>k$: need $D\ge0$, $a\,f(k)>0$, and $-\dfrac{b}{2a}>k$.
• Both roots $<k$: need $D\ge0$, $a\,f(k)>0$, and $-\dfrac{b}{2a}<k$.
• $k$ strictly between the roots: need $a\,f(k)<0$ (this alone also forces $D>0$). [JEE Tip] The sign of $a\,f(k)$ is the fastest discriminator: negative means $k$ is trapped between the roots; positive with $D\ge0$ means both roots lie on the same side of $k$, and the vertex then tells you which side.

7. Roots of a cubic

If $\alpha,\beta,\gamma$ are the roots of $ax^3+bx^2+cx+d=0$, then $$\alpha+\beta+\gamma=-\dfrac{b}{a},\qquad \alpha\beta+\beta\gamma+\gamma\alpha=\dfrac{c}{a},\qquad \alpha\beta\gamma=-\dfrac{d}{a}.$$

Solved Examples — Beyond-NCERT Formulae

Each example names the formula it uses and ends with a quick check.

Example 1 — Rotation: vertices of a square

Problem. $A(2+i)$ and $B(6+4i)$ are two adjacent vertices of a square $ABCD$ labelled anticlockwise. Find the remaining vertices $C$ and $D$.

Solution. Going anticlockwise, the side $\vec{AD}$ is $\vec{AB}$ turned by $+90$ degrees, so $D-A=(B-A)e^{i\pi/2}=(B-A)i$. Here $B-A=(6+4i)-(2+i)=4+3i$, hence $(B-A)i=(4+3i)i=-3+4i$ and $$D=A+(-3+4i)=(2+i)+(-3+4i)=-1+5i.$$ As $ABCD$ is a parallelogram, $C=B+(D-A)=(6+4i)+(-3+4i)=3+8i$.

Check. $|B-A|=|4+3i|=5$ equals $|D-A|=|-3+4i|=5$ (equal sides), and $\operatorname{Re}\big((B-A)\overline{(D-A)}\big)=0$ (adjacent sides perpendicular).

Answer. $C=3+8i$ and $D=-1+5i$.

Example 2 — Rotation: third vertex of an equilateral triangle

Problem. Two vertices of an equilateral triangle are $A(2+2i)$ and $B(6+2i)$. Find the third vertex $C$.

Solution. $C$ comes from rotating $B$ about $A$ through $\pm 60$ degrees: $C=A+(B-A)e^{\pm i\pi/3}$. With $B-A=4$ and $e^{i\pi/3}=\tfrac12+\tfrac{\sqrt3}{2}i$, $$C=A+4\left(\tfrac12+\tfrac{\sqrt3}{2}i\right)=(2+2i)+(2+2\sqrt3\,i)=4+(2+2\sqrt3)i.$$ The $-60$ degree rotation gives the mirror-image apex $C'=4+(2-2\sqrt3)i$.

Check. $|C-A|=|2+2\sqrt3\,i|=\sqrt{4+12}=4=|B-A|$, and $|C-B|=|-2+2\sqrt3\,i|=4$, so all three sides equal $4$.

Answer. $C=4+(2+2\sqrt3)i$ or $C=4+(2-2\sqrt3)i$.

Example 3 — Parallelogram law

Problem. For complex numbers $z_1,z_2$ it is given that $|z_1|=3$, $|z_2|=4$ and $|z_1+z_2|=5$. Find $|z_1-z_2|$.

Solution. By the parallelogram law, $$|z_1+z_2|^2+|z_1-z_2|^2=2\big(|z_1|^2+|z_2|^2\big).$$ So $5^2+|z_1-z_2|^2=2(9+16)=50$, giving $|z_1-z_2|^2=25$ and $|z_1-z_2|=5$.

Check. From $|z_1+z_2|^2=|z_1|^2+|z_2|^2+2\operatorname{Re}(z_1\bar z_2)$ we get $25=25+2\operatorname{Re}(z_1\bar z_2)$, so $\operatorname{Re}(z_1\bar z_2)=0$: the vectors are perpendicular, which is exactly why both diagonals have length $5$.

Answer. $|z_1-z_2|=5$.

Example 4 — Triangle inequality: range of a modulus

Problem. If $|z_1|=3$ and $|z_2|=5$, find the greatest and least possible values of $|z_1+z_2|$.

Solution. The two-sided triangle inequality gives $$\big||z_1|-|z_2|\big|\le |z_1+z_2|\le |z_1|+|z_2|,$$ so $|5-3|\le |z_1+z_2|\le 5+3$, that is $2\le |z_1+z_2|\le 8$. The maximum $8$ is reached when $z_1,z_2$ share the same argument, the minimum $2$ when they are oppositely directed.

Answer. Greatest value $8$, least value $2$.

Example 5 — Exactly one common root

Problem. Using the resultant condition, show that $x^2-5x+6=0$ and $x^2-7x+12=0$ have exactly one common root and find it.

Solution. Take $a_1,b_1,c_1=1,-5,6$ and $a_2,b_2,c_2=1,-7,12$. The one-common-root condition is $(c_1a_2-c_2a_1)^2=(a_1b_2-a_2b_1)(b_1c_2-b_2c_1)$. Compute each block: $$c_1a_2-c_2a_1=6-12=-6,\quad a_1b_2-a_2b_1=-7+5=-2,\quad b_1c_2-b_2c_1=(-5)(12)-(-7)(6)=-18.$$ Then the left side is $(-6)^2=36$ and the right side is $(-2)(-18)=36$; they agree, so exactly one root is shared, and it equals $$x=\dfrac{c_1a_2-c_2a_1}{a_1b_2-a_2b_1}=\dfrac{-6}{-2}=3.$$

Check. $3^2-5(3)+6=0$ and $3^2-7(3)+12=0$. The full root sets are $\{2,3\}$ and $\{3,4\}$, sharing only $3$.

Answer. One common root, $x=3$.

Example 6 — Both roots exceeding a given number

Problem. Find all real $k$ for which both roots of $x^2-6x+k=0$ are greater than $2$.

Solution. Let $f(x)=x^2-6x+k$ with $a=1>0$. Both roots exceed $2$ exactly when three conditions hold together:

• Real roots: $D=36-4k\ge0\Rightarrow k\le 9$.
• Sign at $2$: $a\,f(2)>0\Rightarrow 4-12+k>0\Rightarrow k>8$.
• Vertex right of $2$: $-\dfrac{b}{2a}=3>2$, which holds automatically. Intersecting these gives $8<k\le 9$.

Check. At $k=9$ the equation has the double root $3$ (both $>2$); at $k=\tfrac{17}{2}$ the roots are $3\pm\tfrac{\sqrt2}{2}\approx 2.29,\,3.71$; at $k=8$ one root is exactly $2$, so $k=8$ is excluded.

Answer. $8<k\le 9$.

Example 7 — Products over the $n$th roots of unity

Problem. Let $1,z_1,z_2,z_3,z_4,z_5$ be the six $6$th roots of unity. Find (a) the product of all six roots and (b) $(1-z_1)(1-z_2)(1-z_3)(1-z_4)(1-z_5)$.

Solution. (a) The product of all $n$ roots is $(-1)^{n-1}$; for $n=6$ this is $(-1)^5=-1$. (Equivalently it is the product of the roots of $z^6-1=0$, namely $(-1)^6\cdot(-1)=-1$.) (b) Since $\displaystyle\prod_{k=1}^{5}(x-z_k)=\dfrac{x^6-1}{x-1}=1+x+x^2+x^3+x^4+x^5$, substitute $x=1$: $$(1-z_1)(1-z_2)(1-z_3)(1-z_4)(1-z_5)=1+1+1+1+1+1=6.$$

Answer. (a) $-1$; (b) $6$.

Example 8 — Symmetric functions of a cubic

Problem. If $\alpha,\beta,\gamma$ are the roots of $x^3-6x^2+11x-6=0$, find $\alpha^2+\beta^2+\gamma^2$ and $\dfrac1\alpha+\dfrac1\beta+\dfrac1\gamma$.

Solution. With $a=1,\ b=-6,\ c=11,\ d=-6$, the cubic relations give $$\alpha+\beta+\gamma=6,\qquad \alpha\beta+\beta\gamma+\gamma\alpha=11,\qquad \alpha\beta\gamma=6.$$ Hence $$\alpha^2+\beta^2+\gamma^2=(\alpha+\beta+\gamma)^2-2(\alpha\beta+\beta\gamma+\gamma\alpha)=36-22=14,$$ $$\dfrac1\alpha+\dfrac1\beta+\dfrac1\gamma=\dfrac{\alpha\beta+\beta\gamma+\gamma\alpha}{\alpha\beta\gamma}=\dfrac{11}{6}.$$

Check. The roots are $1,2,3$: indeed $1+4+9=14$ and $1+\tfrac12+\tfrac13=\tfrac{11}{6}$.

Answer. $\alpha^2+\beta^2+\gamma^2=14$ and $\dfrac1\alpha+\dfrac1\beta+\dfrac1\gamma=\dfrac{11}{6}$.