CSIR NET QUESTION Complex Analysis, Real Analysis, and a dash of Algebraic intuition with deep analysis (Round 1)

🔹 Question 1: Complex Numbers – Argument

Let \( z = -1 + i \). What is the principal argument of \( z \)?

• A) \( \frac{3\pi}{4} \)
• B) \( -\frac{\pi}{4} \)
• C) \( \frac{\pi}{4} \)
• D) \( \frac{5\pi}{4} \)

🔹 Complex Numbers – Principal Argument

Given: \( z = -1 + i \)

Since \( z \) lies in the second quadrant, we calculate its argument accordingly:

👉 \( \tan^{-1} \left( \frac{\text{Im}(z)}{\text{Re}(z)} \right) = \tan^{-1} \left( \frac{1}{-1} \right) = \tan^{-1}(-1) \)
The angle corresponding to this is \( -\frac{\pi}{4} \), which lies in the fourth quadrant.
But for the principal argument (second quadrant), we do:
\( =Arg(z) = \pi - \frac{\pi}{4} = \frac{3\pi}{4} \)

✅ Correct Answer: A

Other Options:

• B: \( -\frac{\pi}{4} \) → applicable to 4th quadrant
• C: \( \frac{\pi}{4} \) → 1st quadrant
• D: \( \frac{5\pi}{4} \) → 3rd quadrant

❌ So option A is logically the best fit!

MORE DEEP

🔹 Step-by-Step: Argument of a Complex Number

Definition: Argument of a complex number \( z \) is the angle made by the point \( z \) in the complex plane with the positive x-axis (real axis), measured anticlockwise.

🧭 Step 1: Location of \( z = -1 + i \)

- \( x = -1 \) → Left of origin
- \( y = +1 \) → Above origin
✅ So, the point lies in the Second Quadrant

🎯 Step 2: Calculating the Angle

- Use: \( \tan^{-1} \left( \frac{y}{x} \right) = \tan^{-1} \left( \frac{1}{-1} \right) = \tan^{-1}(-1) \)
- This gives angle \( -\frac{\pi}{4} \), but that's in the Fourth Quadrant
❗ We need the Principal Argument, which must be between \( 0 \) and \( \pi \) in the Second Quadrant

✅ Final Formula

So we use: \( Arg(z) = \pi - \left| -\frac{\pi}{4} \right| = \pi - \frac{\pi}{4} = \frac{3\pi}{4} \)

🔎 Summary of Options

• A: \( \frac{3\pi}{4} \) – ✅ Correct
• B: \( -\frac{\pi}{4} \) – ❌ Fourth Quadrant
• C: \( \frac{\pi}{4} \) – ❌ First Quadrant
• D: \( \frac{5\pi}{4} \) – ❌ Third Quadrant

Final Answer: A

🔹 Question 2: Real Analysis – Continuity

Which of the following functions is discontinuous everywhere?

• A) \( f(x) = x^2 \)
• B) \( f(x) = \sin(1/x),\ f(0) = 0 \)
• C) Dirichlet function: \( f(x) = \begin{cases} 1 & \text{if } x \in \mathbb{Q} \\ 0 & \text{if } x \in \mathbb{R} \setminus \mathbb{Q} \end{cases} \)
• D) \( f(x) = |x| \)

✅ Correct Answer: C (Dirichlet function)

Dirichlet function:
\( f(x) = \begin{cases} 1 & \text{if } x \in \mathbb{Q} \ (\text{rational}) \\ 0 & \text{if } x \in \mathbb{R} \setminus \mathbb{Q} \ (\text{irrational}) \end{cases} \)

🔍 Reason:

• Both rationals and irrationals are dense in every interval of real numbers
• So, in any neighborhood, \( f(x) \) jumps between 1 and 0 repeatedly
• Limit doesn’t exist at any point → discontinuous everywhere

❌ Incorrect Options:

• A) \( f(x) = x^2 \) – Smooth polynomial → continuous everywhere
• B) \( f(x) = \sin(1/x),\ f(0) = 0 \) – Oscillates near 0 → discontinuous only at 0
• D) \( f(x) = |x| \) – “V” shape at origin → still continuous everywhere

🟰 Final Answer: C

🔹 Question 3: Linear Algebra – Diagonalization

Which matrix is diagonalizable?

• A) \( A = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} \)
• B) Identity matrix: \( A = I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \)
• C) Zero matrix: \( A = 0 = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \)
• D) \( A = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix} \)

✅ Correct Answer: B (Identity matrix)

Identity matrix: \( A = I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \)

🔍 Reason:

• Already in diagonal form ✅
• All eigenvalues = 1 ⇒ algebraic multiplicity = geometric multiplicity
• Eigenvectors span the entire space ⇒ fully diagonalizable

❌ Other Options:

• A) \( \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} \) – repeated eigenvalue but lacks enough independent eigenvectors
• C) Zero matrix \( \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \) – diagonalizable, but trivial case
• D) Rotation matrix \( \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix} \) – eigenvalues are imaginary ⇒ not diagonalizable over \(\mathbb{R}\)

🟰 Best choice: B