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