2 Attachment(s)

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

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 $i^9=i$.

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. and .

I don't know what you mean by 'surd9' but I'm sure you did not mean " "!

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

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

Hello, delso!

Multiply by

. .