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4-1.Complex numbers
normal
${(1 + i)^{2n}} = {(1 - i)^{2n}}$ के लिये न्यूनतम धनात्मक पूर्णांक $n$ का मान है
A
$1$
B
$2$
C
$3$
D
$4$
Solution
(b) हम जानते हैं कि ${(1 + i)^{2n}} = {(1 – i)^{2n}}$
$ \Rightarrow {\left( {\frac{{1 + i}}{{1 – i}}} \right)^{2n}} = 1$$ \Rightarrow {(i)^{2n}} = 1$$ \Rightarrow {(i)^{2n}} = {( – 1)^2}$
$ \Rightarrow {(i)^{2n}} = {({i^2})^2}$$ \Rightarrow {(i)^{2n}} = {(i)^4}$
$ \Rightarrow 2n = 4$$ \Rightarrow n = 2$.
Standard 11
Mathematics
