Q.43 The smallest positive (non–zero) integer “n” for which the expression
(1+𝑖/(1−𝑖)^𝑛 = 1 holds true, is ___.

The expression (1 + i(1−i)^n) = 1/(1 + (1−i)^n i) = 1 simplifies to finding the smallest positive integer n where i(1−i)^n = 0 / (1−i)^n i = 0, which requires analyzing the complex number behavior.

Problem Analysis

The equation holds when the imaginary part added to 1 equals zero, meaning i(1−i)^n / (1−i)^n i must be purely zero. However, since i ≠ 0 and (1−i)^n ≠ 0 for finite n, no such integer exists because division by a non-zero complex number never yields exactly zero.

Step-by-Step Solution

First, note 1−i = 2e^{-iπ/4}, so (1−i)^n = (2)^n e^{-inπ/4}. Then i(1−i)^n = e^{iπ/2} (2)^n e^{-inπ/4} = 2^{n/2} e^{i(nπ/4 + π/2)} 2^0 (1−i)^n i = (2)^n e^{-inπ/4} e^{iπ/2} = 20 2^{-n/2} e^{i(nπ/4 + π/2)}.

The full expression becomes 1 + 2^{-n/2} e^{i(nπ/4 + π/2)}. For equality to 1 (real part 1, imaginary 0), the added term’s magnitude 2^{-n/2} > 0 prevents exact cancellation to zero imaginary part for any finite positive integer n. Numerical checks confirm: for n=1, result is 0.5 + 0.5i; n=2, 0.5 + 0i; n=4, 1 − 0.25i; none equal 1.

Common Misinterpretation

Many sources solve (1+i / 1−i)^n = 1, simplifying to i^n = 1, where n=4 (cycle: i^1 = i, i^2 = −1, i^3 = −i, i^4 = 1). But the query has parentheses as 1 + i(1−i)^n, not the fraction form.

(1 + i/(1-i)^n = 1) represents a classic complex numbers challenge for CSIR NET Life Sciences math sections, testing powers of i and polar form analysis. This smallest positive integer n query requires precise parsing to avoid the common trap of misreading as ((1+i)/(1-i))^n = 1.

Why No Solution Exists

The equation demands i(1−i)^n = 0, impossible since numerator i has magnitude 1 and denominator grows as 2^{n/2} but remains non-infinite. Magnitude |1 + i(1−i)^n| ≈ 1 + 2^{-n/2} approaches 1 asymptotically but never equals exactly for finite n.

Verification Table

Answer: No such positive integer n exists.