If 1- i -1- i x -1- thenA- x-4 n- where n is any positive integerB- x-2 n- where n is any positive integerC- x-4 n-1- where n is any positive integerD- x-2 n-1- where n is any positive integer

(1+i/1 i)^x=1⇒[(1+i)^2/1 i^2]^x=1, ⇒[(1+i^2+2i/1+1]^x=1⇒ i^x=1 ∵ x=4n,nℰI^+. ...

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Standard XII

Mathematics

Proof by mathematical induction

If 1+ i /1- i...

Question

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

x = 4n, where n is any positive integer

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x = 2n, where n is any positive integer

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x = 4n + 1, where n is any positive integer

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x = 2n + 1, where n is any positive integer

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Solution

The correct option is A x = 4n, where n is any positive integer
${(\frac{1 + i}{1 - i})}^{x} = 1 \Rightarrow {[\frac{(1 + i)^{2}}{1 - i^{2}}]}^{x} = 1$ ,
$\Rightarrow {[\frac{(1 + i^{2} + 2 i}{1 + 1}]}^{x} = 1 \Rightarrow i^{x} = 1$
$∵ x = 4 n, n E I^{+} .$

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If (1+i1−i)x=1, then

The correct option is A x = 4n, where n is any positive integer(1+i1−i)x=1⇒[(1+i)21−i2]x=1, ⇒[(1+i2+2i1+1]x=1⇒ ix=1 ∵ x=4n,nEI+.

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

The correct option is A x = 4n, where n is any positive integer
${(\frac{1 + i}{1 - i})}^{x} = 1 \Rightarrow {[\frac{(1 + i)^{2}}{1 - i^{2}}]}^{x} = 1$ ,
$\Rightarrow {[\frac{(1 + i^{2} + 2 i}{1 + 1}]}^{x} = 1 \Rightarrow i^{x} = 1$
$∵ x = 4 n, n E I^{+} .$