IIT JEE , AIEEE Forum - Mathematics , Algebra

IITPATIENT

Solve this: 1^i = ? and e^i =?

priyesh

1^i = e^[(i^2)*0)]=e^-0 = 1

e^i = cos1 + isin1

sankarapandian

1^i=1;

e^i=cos1+isin1;

y=1^i=>logy=ilog1=>logy=0;

IITPATIENT

If 1^i=1 then i=infinity but i=(-1)^1/2

(2)

1^i=1;

e^i=cos1+isin1;

y=1^i=>logy=ilog1=>logy=0;

Itfaq2009

1^i= {cos(2pie)+isin(2pie)}^i=(e^i(pie))^i=e^-2pie

ganapathi1810

1=>e power i 2pie

1^i=>e power i^2 2pie

=> e power (-2pie)

e^i => cos1+i sin1

where 1=180degrees\pie=> 180*7\22 degrees