1-i-9-

1-i-9-

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 = \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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