Now, figure e ia
.
e ia = 1 + ia
+(ia) 2/2! + (ia)
3/3!
+ (ia) 4/4! + (ia)
5/5!
+ (ia)
6/6! + ...
Simplify all the powers of i.
= 1 +
ia - a2/2!
- ia3/3! + a4/4!
+ ia5/5! - a6/6!
+ ...
Regroup:
= 1 -
a2/2!
+ a4/4! - a6/6!
+ ia
- ia3/3!
+ ia5/5! - ia7/7!...
= (1 -
a2/2!
+ a4/4! - a6/6!+...)
+ i(a
- a3/3!
+ a5/5! - a7/7!...)
Substitute in the above series names:
= cosa
+ i sin a
So
e ia
= cos a + i sin a
= cis a
Example: 1 - i = \/2 cis (-p/4) = ei (p/4)
Now, this means eip = cos
p
+ i sin p = -1 + 0i = -1
So, eip = -1.
Also, if eip = -1, then ln (-1) = ip and we can actually talk about logs of negative numbers.
Example: ln (-45.67) = ln ((-1)(45.67) = ln (-1)
+ ln 45.67 = p + 3.821
or approximately 6.935
Problems (there are no problems here yet)
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