If ( 1 − i 1 + i ) x = 1 then

( 1 − i 1 + i ) x = 1 ⇒ [ 1 − i 2 ( 1 + i ) ] x = 1 ( 1 + 1 1 + i 2 + 2 i ) x = 1 ⇒ ( i ) x = 1 ; ∴ x = 4 n ; n ∈ 1 +

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MathematicsComplex NumberGeneral(Modulus,Argument,Conjugate)Medium2 minPYQ_2003

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If $(\frac{1 + i}{1 - i})^{x} = 1$ then

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Solution:

$(\frac{1 + i}{1 - i})^{x} = 1 \Rightarrow [\frac{( 1 + i )}{1 - i ^{2}}]^{x} = 1$ $(\frac{1 + i ^{2} + 2 i}{1 + 1})^{x} = 1 \Rightarrow (i)^{x} = 1; ∴ x = 4 n; n \in 1^{+}$

Stream:JEESubject:MathematicsTopic:Complex NumberSubtopic:General(Modulus,Argument,Conjugate)

⏱ 2mℹ️ Source: PYQ_2003

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