Complex Numbers: Conceptual Doubts Cleared (8)

Question

If $z = \dfrac{1 + i}{1 - i}$, find $z^{100}$ and $\arg(z^{100})$. JEE Main 2024 standard.

Solution — Step by Step

Why This Works

The trick of multiplying by the conjugate clears the imaginary part from the denominator. After simplification, $z$ collapses to a recognisable value ($i$ in this case), and the rest is just exploiting the cyclic structure of powers of $i$.

The key insight: powers of $i$ have period 4. To compute $i^n$, just find $n \mod 4$ and read off from $\{1, i, -1, -i\}$.

Alternative Method

Convert $z$ to polar form: $z = i = e^{i\pi/2}$, so $|z| = 1$ and $\arg(z) = \pi/2$.

$z^{100} = e^{i \cdot 100\pi/2} = e^{i \cdot 50\pi}$. Since $e^{i\theta}$ has period $2\pi$, reduce $50\pi \mod 2\pi$: $50\pi = 25 \cdot 2\pi$, which is $0 \mod 2\pi$. So $z^{100} = e^{i0} = 1$. Same answer.

This polar approach generalises to non-trivial $|z|$ values where $i^n$ tricks don’t apply.