(1+i)² (1-i²) simplify
Answers
Correct Question :
Simplify (1 + i)² + (1 - i)²
Solution :
=> (1 + i)² + (1 - i)²
we need that,
- (a + b)² = a² + b² + 2ab
- (a - b)² = a² + b² - 2ab
=> 1² + i² + 2i + 1² + i² - 2i
=> 1 + i² + 1 + i²
=> 2 + 2i²
- i² = -1
=> 2 + 2 × (-1)
=> 2 - 2
=> 0
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Step-by-step explanation:
The standard approach when dealing with the addition or subtraction of fractions with different denominators is to arithmetically manipulate the fractions to have a common denominator.
In your example, the common denominator we require is the lowest common multiple of the two fractions’ denominators, which is (1-i)²(1+i)³
The expression thus becomes:
(1+i)6−(1−i)4(1−i)2(1+i)3(1+i)6−(1−i)4(1−i)2(1+i)3
We could then use the binomial theorem to expand these terms, which would give us a polynomial of degree 6 over a polynomial of degree 5.
However, assuming that i=−1−−−√i=−1 , the fraction would eventually simplify to:
a+ibc+ida+ibc+id
We can then remove an imaginary component from the denominator by multiplying both parts of the fraction by (c−id)(c−id) , which gives us:
(a+ib)(c−id)(c+id)(c−id)(a+ib)(c−id)(c+id)(c−id)
=(ac−bd)+(bc−ad)ic2+d2=(ac−bd)+(bc−ad)ic2+d2
Sure, we can work through all of this, but it all seems a tad tedious