Find the integral solutions of the equations
${(1 + i)}^{n} = {(1 - i)}^{n} .$

n = 2 k

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n = 3 k

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n = 4 k

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n = 2 k + 1

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Solution

The correct option is C $n = 4 k$
Given, $(1 + i)^{n} = (1 - i)^{n}$

$(\sqrt{2})^{n} (\frac{1 + i}{\sqrt{2}})^{n} = (\sqrt{2})^{n} (\frac{1 - i}{\sqrt{2}})^{n}$

$∴ e^{\frac{i n π}{4}} = e^{\frac{- n i π}{4}}$

$∴ e^{\frac{n i π}{2}} = 1$

$∴ e^{\frac{n i π}{2}} = e^{i 2 k π}$

$∴ \frac{n π}{2} = 2 k π$

$∴ \frac{n}{2} = 2 k$

$∴ n = 4 k$ .

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Find the integral solutions of the equations(1+i)n=(1−i)n.

The correct option is C n=4kGiven, (1+i)n=(1−i)n(√2)n(1+i√2)n=(√2)n(1−i√2)n∴einπ4=e−niπ4∴eniπ2=1∴eniπ2=ei2kπ∴nπ2=2kπ∴n2=2k∴n=4k.

Find the integral solutions of the equations
${(1 + i)}^{n} = {(1 - i)}^{n} .$