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Algebra

Cool goIITian

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31 Aug 2008 23:19:17 IST

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

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

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Priyesh

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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

sankar

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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;

Naveen kumar

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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

Amey9420651744

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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;

Sandeep

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Joined: 1 Sep 2008

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1 Sep 2008 17:19:46 IST

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

ganapathi 1810

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Joined: 28 Aug 2008

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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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