Start learning →

Indian Navy SSR Complex Numbers and Quadratic Equations — practice questions

Explore 5 free Complex Numbers and Quadratic Equations questions in Indian Navy SSR Mathematics. Open each question or exercise to read its content, reference answers and available explanation.

Practise Complex Numbers and Quadratic Equations with Quill →

Complex Numbers and Quadratic Equations questions and exercises with answers

Showing questions 1–5 of 5.

What is the principal argument of the complex number $z = 1 + i$?Read the answer and worked solution →The locus of the point $z$ in the Argand plane satisfying $|z - 2| = |z + 2|$ is:Read the answer and worked solution →The imaginary unit i is defined as:Read the answer and worked solution →The complex conjugate of $3 - 4i$ is:Read the answer and worked solution →The modulus $|1 - i|$ equals:Read the answer and worked solution →

Worked practice questions

What is the principal argument of the complex number $z = 1 + i$?

  1. $\dfrac{\pi}{3}$
  2. $\dfrac{\pi}{4}$
  3. $\dfrac{\pi}{6}$
  4. $\dfrac{\pi}{2}$
Show answer and solution

Correct answer: B. $\dfrac{\pi}{4}$

$\tan\theta = \dfrac{y}{x} = 1$, and $z$ is in the first quadrant, so $\theta = \dfrac{\pi}{4}$.

Open this question →

The locus of the point $z$ in the Argand plane satisfying $|z - 2| = |z + 2|$ is:

  1. A parabola
  2. A circle of radius $2$ centred at the origin
  3. The real axis ($\mathrm{Im}(z) = 0$)
  4. The imaginary axis ($\mathrm{Re}(z) = 0$)
Show answer and solution

Correct answer: D. The imaginary axis ($\mathrm{Re}(z) = 0$)

**Geometric reading.** $|z - 2|$ is the distance from the point $z$ to $(2, 0)$. $|z + 2|$ is the distance to $(-2, 0)$. Equating them means $z$ is equidistant from the two points. **Locus.** Set of points equidistant from $(2, 0)$ and $(-2, 0)$ is the **perpendicular bisector** of the segment joining them — the $y$-axis, i.e. $\mathrm{Re}(z) = 0$. **Algebraic confirmation.** Let $z = x + iy$. Squaring: $(x-2)^2 + y^2 = (x+2)^2 + y^2 \Rightarrow -4x = 4x \Rightarrow x = 0$. **Why option C is tempting.** A symmetric-looking radius-2 hint, but $|z - 2|$ is *not* the same as $|z| - 2$ it's a distance to a *point*, not a centred circle equation.

Open this question →

The imaginary unit i is defined as:

  1. √(-1)
  2. √1
  3. 1/i
  4. -1
Show answer and solution

Correct answer: A. √(-1)

i = √(-1) by definition. **So i² = -1. This is the foundation of complex numbers, allowing extension beyond real numbers.**

Open this question →

How to practise

  1. Choose a Complex Numbers and Quadratic Equations question and try to work out your answer.
  2. Open the question to check the correct answer and read its worked solution.
  3. Continue in the app for bite-sized lessons, explanations and review with Quill. You can start without creating an account.

More Mathematics topics

SequencesBinomial TheoremSequences and SeriesProbability & StatisticsGeometryPermutations and CombinationsLinear ProgrammingVectors & 3D Geometry