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Miscellaneous Exercise · Q14

Q.If (1+i1−i)m=1\left(\dfrac{1 + i}{1 - i}\right)^{m} = 1, then find the least positive integral value of mm.

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The key idea is to simplify the complex fraction 1+i1−i\frac{1+i}{1-i} to ii, then find the smallest positive integer mm such that im=1i^m = 1. The least positive integral value of mm is 44.

This problem is a classic example of how a complex number expression can be tamed by first simplifying its core. Instead of wrestling with powers of a messy fraction, we reduce the base to its simplest a+iba+ib form and then work with powers of that simple result.

The expression 1+i1−i\frac{1+i}{1-i} looks like a division of two complex numbers. The standard trick is to multiply numerator and denominator by the conjugate of the denominator, which makes the denominator real and lets the fraction collapse to a simple value.

Let's verify this step by step.

  1. Simplify the base fraction. Multiply numerator and denominator by the conjugate of the denominator, 1+i1+i:

1+i1−i⋅1+i1+i=(1+i)212−i2\frac{1+i}{1-i} \cdot \frac{1+i}{1+i} = \frac{(1+i)^2}{1^2 - i^2}

Since i2=−1i^2 = -1, the denominator becomes 1−(−1)=21 - (-1) = 2. The numerator expands:

(1+i)2=1+2i+i2=1+2i−1=2i(1+i)^2 = 1 + 2i + i^2 = 1 + 2i - 1 = 2i

So the whole fraction simplifies to:

2i2=i\frac{2i}{2} = i

Tip

Notice both 1+i1+i and 1−i1-i have the same modulus (2\sqrt{2}), so the quotient has modulus 11 — this is why the simplified fraction turned out to be a number like ii that lies exactly on the unit circle.

  1. Rewrite the given equation. The original condition becomes: …

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