[QUOTE=delso;821295]i express (1+i)^9 as 16surd9(cos (9pi/4) +i sin (9pi/4)) and (1-i)^9 as 16surd9(cos (-9pi/4) -i sin (9pi/4))[quote]
This is wrong. $\displaystyle |1+ i|= \sqrt{2}$ and $\displaystyle (\sqrt{2})^9= \sqrt{2}(\sqrt{2})^8= \sqrt{2}(2)^4= 16\sqrt{2}$.
I don't know what you mean by 'surd9' but I'm sure you did not mean "$\displaystyle \sqrt{2}$"!
, then i divide both, but my ans is (1+i)/(1-i) , which is differnt with the ans given, why cant i do in this way? i have attached the sample ans
Hello, delso!
$\displaystyle \text{(b) Evaluate: }\:\left(\frac{1+i}{1-i}\right)^9$
Multiply by $\displaystyle \frac{1+i}{1+i}$
. . $\displaystyle \left(\frac{1+i}{1-i}\cdot{\color{red}\frac{1+i}{1+i}}\right)^9 \;=\;\left(\frac{1+2i-1}{1-(\text{-}1)}\right)^9 \;=\;\left(\frac{2i}{2}\right)^9 \;=\;i^9 \;=\;i $