Let S = z : 3 ≤ |2z - 3(1 + i)| ≤ 7$S = \{z : 3 \leq |2z - 3(1 + i)| \leq 7\}$ be a set of complex numbers. Then z in S | ( z + (1)/(2)(5 + 3i) ) |$\min_{z \in S} \left| \left( z + \frac{1}{2}(5 + 3i) \right) \right|$ is equal to:
This represents an annular region bounded by two concentric circles centered at C = (3)/(2) + (3)/(2)i$C = \frac{3}{2} + \frac{3}{2}i$ with radii r₁ = (3)/(2)$r_1 = \frac{3}{2}$ and r₂ = (7)/(2)$r_2 = \frac{7}{2}$.
Geometry of Complex Numbers diagram for Q17 - JEE Main 2026 Morning
Step 1: Identify the Target Point
We need to minimize | z - ( -(5)/(2) - (3)/(2)i ) |$\left| z - \left( -\frac{5}{2} - \frac{3}{2}i \right) \right|$.
Let P$P$ be the point -(5)/(2) - (3)/(2)i$-\frac{5}{2} - \frac{3}{2}i$. The expression represents the distance from point P$P$ to a point z$z$ in the set S$S$.
Step 2: Distance from Center to P
Calculate the distance PC$PC$ between the center of the circles C((3)/(2), (3)/(2))$C\left(\frac{3}{2}, \frac{3}{2}\right)$ and the point P(-(5)/(2), -(3)/(2))$P\left(-\frac{5}{2}, -\frac{3}{2}\right)$:
Since 5 > (7)/(2)$5 > \frac{7}{2}$, the point P$P$ lies outside the outer circle.
Step 3: Minimum Distance Calculation
The shortest distance from an external point to an annular region is the distance to the outer boundary along the line connecting the point to the center.
z in S |z - P| = PC - router$$\min_{z \in S} |z - P| = PC - r_{\text{outer}}$$Minimum Distance = 5 - (7)/(2) = (10 - 7)/(2) = (3)/(2)$$\text{Minimum Distance} = 5 - \frac{7}{2} = \frac{10 - 7}{2} = \frac{3}{2}$$
Pattern Recognition
Transforming |az - b|$|az - b|$ by factoring out a$a$ immediately reveals the true geometric center. Shortest distance to any ring/circle from an external point is always collinear with the center: d - Router$d - R_{outer}$.
Chapter Mix
Class 11 Maths: Complex Numbers and Quadratic Equations
Keywords:#minimum distance complex numbers#JEE Main 2026 Morning Q17#Complex Numbers and Quadratic Equations JEE Main 2026#Geometry of Complex Numbers JEE Main 2026
More Complex Numbers and Quadratic Equations Previous-Year Questions — Page 2
Convert standard coordinate complex numbers into Euler form immediately when large powers are present.
Chapter Mix
Class 11 Maths: Complex Numbers
Q9jee_main_2026_24_january_morningLocus in Complex Plane
Let S = z in C : | (z - 6i)/(z - 2i) | = 1 and | (z - 8 + 2i)/(z + 2i) | = (3)/(5)$S = \left\{ z \in \mathbb{C} : \left| \frac{z - 6i}{z - 2i} \right| = 1 \text{ and } \left| \frac{z - 8 + 2i}{z + 2i} \right| = \frac{3}{5} \right\}$. Then Σz in S |z|²$\sum_{z \in S} |z|^2$ is equal to
A.398$398$
B.413$413$
C.423$423$
D.385$385$
Solution
Related Formula
|z - z₁| = |z - z₂| represents the perpendicular bisector of the segment joining z₁ and z₂$$|z - z_1| = |z - z_2| \text{ represents the perpendicular bisector of the segment joining } z_1 \text{ and } z_2$$|x+iy|² = x² + y²$$|x+iy|^2 = x^2 + y^2$$
Core Logic
First condition: |z - 6i| = |z - 2i|$|z - 6i| = |z - 2i|$.
This means z$z$ lies on the perpendicular bisector of (0,6)$(0,6)$ and (0,2)$(0,2)$.
Let z = x + iy$z = x + iy$. Thus, y = 4$y = 4$.
Step 1: Circle Equation
Second condition: 5|z - 8 + 2i| = 3|z + 2i|$5|z - 8 + 2i| = 3|z + 2i|$.
Substitute y = 4$y = 4$ into z$z$: z = x + 4i$z = x + 4i$.
Whenever an absolute value ratio equals 1, immediately map it to a line (perpendicular bisector) and substitute its constraint directly into the second curve equation to reduce dimensionality.
Chapter Mix
Class 11 Maths: Complex Numbers and Quadratic Equations
Q23jee_main_2026_24_january_eveningProperties of Moduli
Let z = (1 + i)(1 + 2i)(1 + 3i) (1 + ni)$z = (1 + \mathrm{i})(1 + 2\mathrm{i})(1 + 3\mathrm{i}) \dots (1 + \mathrm{ni})$, where i = √(-1)$\mathrm{i} = \sqrt{-1}$. If |z|² = 44200$|z|^2 = 44200$, then n$n$ is equal to
The calculated product for n=5$n=5$ exactly matches the prime factorization of 44200$44200$.
Therefore, n = 5$n = 5$.
Pattern Recognition
Modulus is multiplicative. In problems featuring chains of complex multiplications set equal to a huge real magnitude, instantly switch to magnitudes and map to integer factorization.
Chapter Mix
Class 11 Maths: Complex Numbers
Q5jee_main_2026_28_january_morningGeometry of Complex Numbers
Let z$z$ be a complex number such that |z - 6| = 5$|z - 6| = 5$ and |z + 2 - 6i| = 5$|z + 2 - 6i| = 5$. Then the value of z³ + 3z² - 15z + 141$z^{3} + 3z^{2} - 15z + 141$ is equal to
A.42$42$
B.37$37$
C.50$50$
D.61$61$
Solution
Core Logic
Geometry of Complex Numbers
The given equations represent two circles in the complex plane:
Circle 1: Center C₁(6, 0)$C_1(6, 0)$, radius r₁ = 5$r_1 = 5$
Circle 2: Center C₂(-2, 6)$C_2(-2, 6)$, radius r₂ = 5$r_2 = 5$
When given two complex distance modulus equations |z-z₁|=r₁$|z-z_1|=r_1$ and |z-z₂|=r₂$|z-z_2|=r_2$, always check the distance between centers |z₁ - z₂|$|z_1 - z_2|$. If it exactly equals r₁ + r₂$r_1 + r_2$, the single unique solution is the section formula midpoint.
Chapter Mix
Class 11 Mathematics: Complex Numbers and Quadratic Equations
Q10jee_main_2026_28_january_morningNature of Roots
If α, β$\alpha, \beta$, where α < β$\alpha < \beta$, are the roots of the equation λ x² - (λ + 3)x + 3 = 0$\lambda x^2 - (\lambda + 3)x + 3 = 0$ such that (1)/(α) - (1)/(β) = (1)/(3)$\frac{1}{\alpha} - \frac{1}{\beta} = \frac{1}{3}$, then the sum of all possible values of λ$\lambda$ is:
A.6$6$
B.2$2$
C.4$4$
D.8$8$
Solution
Related Formula
For a quadratic equation ax² + bx + c = 0$ax^2 + bx + c = 0$:
Sum of roots: α + β = -(b)/(a)$\alpha + \beta = -\frac{b}{a}$
Product of roots: αβ = (c)/(a)$\alpha\beta = \frac{c}{a}$
Core Logic
From the given equation λ x² - (λ + 3)x + 3 = 0$\lambda x^2 - (\lambda + 3)x + 3 = 0$:
Practice past-year questions one chapter at a time. Pick an exam → subject → chapter and get every PYQ for that topic — pulled together from all past papers — with the chapter's key formulas alongside.
