# complex numer express (1+i)^9/(1-i)^9

• Jun 3rd 2014, 05:49 AM
delso
complex numer express (1+i)^9/(1-i)^9
Attachment 31061Attachment 31060i 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)) , 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
• Jun 3rd 2014, 06:13 AM
Plato
Re: complex numer express (1+i)^9/(1-i)^9
Quote:

Originally Posted by delso
why cant i do in this way? i have attached the sample ans

Whoever wrote that did not know much about the fundamentals.

Here is a very useful fact: $\dfrac{1}{z} = \dfrac{{\overline z }}{{{{\left| z \right|}^2}}}$

Thus $\dfrac{{1 + i}}{{1 - i}} = \dfrac{{{{(1 + i)}^2}}}{2} = i$ and $i^9=i$.
• Jun 3rd 2014, 06:28 AM
HallsofIvy
Re: complex numer express (1+i)^9/(1-i)^9
[QUOTE=delso;821295]Attachment 31061Attachment 31060i 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 "$\displaystyle \sqrt{2}$"!

Quote:

, 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
• Jun 3rd 2014, 07:44 AM
Soroban
Re: complex numer express (1+i)^9/(1-i)^9
Hello, delso!

Quote:

$\displaystyle \text{(b) Evaluate: }\:\left(\frac{1+i}{1-i}\right)^9$

Multiply by $\displaystyle \frac{1+i}{1+i}$

. . $\displaystyle \left(\frac{1+i}{1-i}\cdot{\color{red}\frac{1+i}{1+i}}\right)^9 \;=\;\left(\frac{1+2i-1}{1-(\text{-}1)}\right)^9 \;=\;\left(\frac{2i}{2}\right)^9 \;=\;i^9 \;=\;i$