Question:

\(If (\frac{(1+i}{1-i})^m=1, \text{then find the least positive integral value of m.}\)

Updated On: Feb 10, 2024
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Solution and Explanation

\((\frac{(1+i}{1-i})^m=1\)

\(⇒(\frac{1+i}{1-i}×\frac{1+i}{1+i})^m=1\)

\(⇒(\frac{(1+i)^2}{1^2-1^2})^m=1\)

\(⇒(\frac{1^2+i^2+2i}{2})^m=1\)

\(⇒(\frac{1-1+2i}{2})^m=1\)

\(⇒(\frac{2i}{2})^m=1\)

\(⇒i^m=1\)

\(\text{∴\,m=4k, where k is some integer.}\)

\(\text{Therefore, the least positive integer is 1. }\)

\(\text{Thus, the least positive integral value of m is 4 (= 4 × 1). }\)

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Concepts Used:

Complex Number

A Complex Number is written in the form

a + ib

where,

  • “a” is a real number
  • “b” is an imaginary number

The Complex Number consists of a symbol “i” which satisfies the condition i^2 = −1. Complex Numbers are mentioned as the extension of one-dimensional number lines. In a complex plane, a Complex Number indicated as a + bi is usually represented in the form of the point (a, b). We have to pay attention that a Complex Number with absolutely no real part, such as – i, -5i, etc, is called purely imaginary. Also, a Complex Number with perfectly no imaginary part is known as a real number.