Question 20 If [(1+i) / (1-i)]^m = 1, then find the least positive integral value of m.
Class X1 - Maths -Complex Numbers and Quadratic Equations Page 113
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106
{(1 + i)/(1 - i)}^m = 1
multiply (1 + i) numberator as well as denominator .
{(1 + i)(1 + i)/(1 - i)(1 + i)}^m = 1
{(1 + i)²/(1² - (i)²)}^m = 1
{(1 + i² +2i)/2 }^m = 1
{( 2i)/2}^m = 1
{i}^m = 1
we know, i^4n = 1 where , n is an integer.
so, m = 4n where n is an integers
e.g m = 4 { because least positive integer 1 }
hence, m = 4 ( answer )
multiply (1 + i) numberator as well as denominator .
{(1 + i)(1 + i)/(1 - i)(1 + i)}^m = 1
{(1 + i)²/(1² - (i)²)}^m = 1
{(1 + i² +2i)/2 }^m = 1
{( 2i)/2}^m = 1
{i}^m = 1
we know, i^4n = 1 where , n is an integer.
so, m = 4n where n is an integers
e.g m = 4 { because least positive integer 1 }
hence, m = 4 ( answer )
Answered by
4
{(1 + i)/(1 - i)}^m = 1
multiply (1 + i) numberator as well as denominator .
{(1 + i)(1 + i)/(1 - i)(1 + i)}^m = 1
{(1 + i)²/(1² - (i)²)}^m = 1
{(1 + i² +2i)/2 }^m = 1
{( 2i)/2}^m = 1
{i}^m = 1
we know, i^4n = 1 where , n is an integer.
so, m = 4n where n is an integers
e.g m = 4 { because least positive integer 1 }
hence, m = 4 ( answer )
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