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

n = 1

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n = 2

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n = 3

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n = 4

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Solution

The correct option is D $n = 4$
$(1 + i)^{n} = (1 - i)^{n}$
$\Rightarrow {(\frac{1 + i}{1 - i})}^{n} = 1$
$\Rightarrow {[\frac{{(1 + i)}^{2}}{(1 - i) (1 + i)}]}^{n} = 1$
$\Rightarrow {(\frac{1 - 1 + 2 i}{1 + 1})}^{n} = 1$
$\Rightarrow i^{n} = 1$
Therefore, $n = 4$
Ans: D

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Find the integral solution of (1+i)n=(1−i)n

The correct option is D n=4(1+i)n=(1−i)n⇒(1+i1−i)n=1⇒[(1+i)2(1−i)(1+i)]n=1⇒(1−1+2i1+1)n=1⇒in=1Therefore, n=4Ans: D

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

The correct option is D $n = 4$
$(1 + i)^{n} = (1 - i)^{n}$
$\Rightarrow {(\frac{1 + i}{1 - i})}^{n} = 1$
$\Rightarrow {[\frac{{(1 + i)}^{2}}{(1 - i) (1 + i)}]}^{n} = 1$
$\Rightarrow {(\frac{1 - 1 + 2 i}{1 + 1})}^{n} = 1$
$\Rightarrow i^{n} = 1$
Therefore, $n = 4$
Ans: D