Difficulty distribution
How the classified questions are distributed by difficulty.
Your cart is empty.
Practice Complex Numbers - Algebra - Mathematics previous year questions organised from real papers, with year-wise coverage and clear topic navigation.
Every graph below is calculated only from this selection.
Year-wise coverage for Complex Numbers. Each bar uses a separate theme-derived color.
How the classified questions are distributed by difficulty.
MCQ, numerical, multiple-select and other formats found in these papers.
Top subjects by unique question coverage.
Top topics across the included previous year papers.
Top subtopics inside this exact selection.
Question coverage for the most populated papers. Every active PYP paper remains listed below.
Newest papers appear first. Sort by year, question coverage or name.
| Paper | Year / session | Questions in this view | Open |
|---|---|---|---|
| BITSAT 2025 | 2025 | 2 | View paper |
| BITSAT 2024 | 2024 | 2 | View paper |
| BITSAT 2023 | 2023 | 2 | View paper |
| BITSAT 2022 | 2022 | 2 | View paper |
| BITSAT 2021 | 2021 | 1 | View paper |
| BITSAT 2020 | 2020 | 3 | View paper |
Practice every matching question in batches of 20, with every available option.
If \(z = r{e^{i\theta }}\), then arg(eiz) is
If \(z = {{7 + i} \over {3 + 4i}}\), then z14 is
The root of the equation \(2(1 + i){x^2} - 4(2 - i)x - 5 - 3i = 0\), where \(i = \sqrt { - 1}\), which has greater modulus, is
If Re(z + 2) = | z \(-\) 2 |, then the locus of z is
The smallest positive integral value of n such that \({\left[ {{{1 + \sin {\pi \over 8} + i\cos {\pi \over 8}} \over {1 + \sin {\pi \over 8} - i\cos {\pi \over 8}}}} \right]^n}\) is purely imaginary, is equal to
If \(|w| = 2\), then the set of points \(z = w - {1 \over w}\) is contained in or equal to the set of points z satisfying
Number of solutions of the equation \(z^2+|z|^2=0\) and \(z \neq 0\) is
If \(z_1\) and \(z_2\) be nth root of unity which subtend a right angled at the origin. Then, \(n\) must be of the form
Let $z$ be a complex number for which $\left|2 z \cos \theta+z^2\right|>1$, if $|z|
If ' $a$ ' is a complex number such that $|a|=1$. Find the value of $a$, so that the equation $a z^2+z+1=0$ has one purely imaginary root.