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![[Post New]](/templates/default/images/icon_minipost_new.gif) 31 Aug 2008 23:19:17 IST
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Solve this: 1^i = ? and e^i =?
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1 Sep 2008 00:04:22 IST
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1^i = e^[(i^2)*0)]=e^-0 = 1
e^i = cos1 + isin1
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1 Sep 2008 00:10:51 IST
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1^i=1;
e^i=cos1+isin1;
y=1^i=>logy=ilog1=>logy=0;
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1 Sep 2008 00:22:33 IST
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If 1^i=1 then i=infinity but i=(-1)^1/2
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1 Sep 2008 13:37:24 IST
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1^i=1;
e^i=cos1+isin1;
y=1^i=>logy=ilog1=>logy=0;
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1^i= {cos(2pie)+isin(2pie)}^i=(e^i(pie))^i=e^-2pie
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1 Sep 2008 19:35:09 IST
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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
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