JAMIA 2026 — Mathematics PYQ

To solve this problem, we need to understand the behavior of the imaginary unit $i$, where $i=\sqrt{-1}$.

The powers of $i$ repeat in a cycle of 4:

• $i^1=i$

• $i^2=-1$

• $i^3=-i$

• $i^4=1$

Step 1: Simplify $i^{13}$

To find higher powers of $i$, we divide the exponent by 4 and look at the remainder:

$$13=(4\times 3)+1$$

Since the remainder is 1:

$$i^{13}=i^1=i$$

Step 2: Simplify ${\left(\frac{1}{i}\right)}^{13}$

First, let's simplify the term $\frac{1}{i}$ by multiplying the numerator and denominator by $i$:

$$\frac{1}{i}=\frac{1\times i}{i\times i}=\frac{i}{i^2}=\frac{i}{-1}=-i$$

Now, substitute this back into the power expression:

$${\left(\frac{1}{i}\right)}^{13}=(-i{)}^{13}$$

Since the exponent 13 is an odd number, the negative sign stays outside:

$$(-i{)}^{13}=-(i^{13})$$

From Step 1, we already know that $i^{13}=i$. Therefore:

$${\left(\frac{1}{i}\right)}^{13}=-i$$

Step 3: Substitute values into the main expression

Now substitute the simplified components back into the original equation:

$${\left(i^{13}-{\left(\frac{1}{i}\right)}^{13}\right)}^2=(i-(-i){)}^2$$

Simplify inside the parenthesis:

$$(i+i{)}^2=(2i{)}^2$$

Expand the square:

$$(2i{)}^2=4i^2$$

Since $i^2=-1$:

$$4(-1)=-4$$

Correct Answer:

(b) -4