(1+i)^9 =?


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Question Number 122904 by Khalmohmmad last updated on 20/Nov/20

${(1 + i)}^{9} = ?$
	Answered by Bird last updated on 20/Nov/20
	$1+i=(√2)e^((iπ)/4) ⇒(1+i)^9 =((√2))^9 e^(((k9π)/4)i) =2^(9/2) {cos(((9π)/4))+isin(((9π)/4))} cos(((9π)/4))=cos(2π+(π/4))=cos((π/4))=(1/( (√2))) sin(((9π)/4))=sin(2π+(π/4))=sin((π/4))=(1/( (√2))) ⇒(1+i)^(9 ) =(2^(9/2) /2^(1/2) )(1+i) =2^4 (1+i)=16(1+i)$
	$1 + i = \sqrt{2} e^{\frac{i π}{4}} \Rightarrow {(1 + i)}^{9} = {(\sqrt{2})}^{9} e^{\frac{k 9 π}{4} i}$ $= 2^{\frac{9}{2}} {c o s (\frac{9 π}{4}) + i s i n (\frac{9 π}{4})}$ $c o s (\frac{9 π}{4}) = c o s (2 π + \frac{π}{4}) = c o s (\frac{π}{4}) = \frac{1}{\sqrt{2}}$ $s i n (\frac{9 π}{4}) = s i n (2 π + \frac{π}{4}) = s i n (\frac{π}{4}) = \frac{1}{\sqrt{2}}$ $\Rightarrow {(1 + i)}^{9} = \frac{2^{\frac{9}{2}}}{2^{\frac{1}{2}}} (1 + i)$ $= 2^{4} (1 + i) = 16 (1 + i)$
	Answered by Olaf last updated on 20/Nov/20

	${(1 + i)}^{9} = {[\sqrt{2} e^{i \frac{π}{4}}]}^{9} = 2^{\frac{9}{2}} e^{i \frac{9 π}{4}}$ $= 16 \sqrt{2} e^{i (\frac{π}{4} + 2 π)} = 16 \sqrt{2} (\frac{1 + i}{\sqrt{2}}) = 16 (1 + i)$
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