If 1-i-1- i 100- a - ib- find a - b

(1 i/1+i )^100=a+ib ⇒ ((1 i)(1 i)/(1+i)(1 i) )^100=a+ib [Rationalizing the denominator] ⇒ ((1 2i 1)/(1+1) )^100=a+ib ⇒ ( 2i/2 )^100=a+ib ⇒ ( i)^100=a+ib ⇒ 1= ...

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Byju's Answer

Standard XII

Mathematics

Equality of 2 Complex Numbers

If 1-i/1+ i 1...

Question

$If {(\frac{1 - i}{1 + i})}^{100} = a + i b, find (a, b) .$

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Solution

${(\frac{1 - i}{1 + i})}^{100} = a + i b \Rightarrow {(\frac{(1 - i) (1 - i)}{(1 + i) (1 - i)})}^{100} = a + i b [Rationalizing the denominator] \Rightarrow {(\frac{(1 - 2 i - 1)}{(1 + 1)})}^{100} = a + i b \Rightarrow {(\frac{- 2 i}{2})}^{100} = a + i b \Rightarrow (- i)^{100} = a + i b \Rightarrow 1 = a + i b$

Comparing, we get (a. b) = (1, 0)

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If (1−i1+i)100=a+ib, find (a, b).

(1−i1+i)100=a+ib⇒ ((1−i)(1−i)(1+i)(1−i))100=a+ib [Rationalizing the denominator]⇒ ((1−2i−1)(1+1))100=a+ib⇒ (−2i2)100=a+ib⇒ (−i)100=a+ib⇒ 1=a+ib Comparing, we get (a. b) = (1, 0)

$If {(\frac{1 - i}{1 + i})}^{100} = a + i b, find (a, b) .$