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Find the value

posted by Ragib
- March 07 2011 07:54:19 AM

(-1)^i=?.i=imaginery number=root over (-1)

Reply by Caprico
- March 10 2011 10:28:13 AM

0.04321

I wonder why you ask this. Its a high-level mathematical problem. Took me quite some time. Had to convert to Euler's form and then proceed. Unless you have a trick or shortcut, I feel this riddle is just a maths problem.

I wonder why you ask this. Its a high-level mathematical problem. Took me quite some time. Had to convert to Euler's form and then proceed. Unless you have a trick or shortcut, I feel this riddle is just a maths problem.

Reply by Ragib
- March 11 2011 09:54:46 AM

you don't even need to think if you wanna cheat.google it and its the first result.

Reply by Ragib
- March 11 2011 10:32:26 AM

did you google it?it's very hard to get to eulers identity without knowing it well.what's the euler form of your answer?

Reply by Caprico
- March 14 2011 01:38:58 AM

Hey, i didn't Google it or cheat. I'm 17 years old, and coaching for IIT. This has been taught to us. I used a scientific calculator for the last calculation. That's how I got this number. I know complex numbers very well. Its one of my favorite chapters. I'll explain my solution in the next comment.

Reply by Caprico
- March 14 2011 01:47:21 AM

(-1)^i = (i^2)^i = i^(2i) = (i^i)^2

Now the main question becomes i^i. The base i can be written as e^iθ. This makes it

e^[(i^2)θ]

= e^-θ

According to what we've been taught, the general value for θ is kπ + π/2. The principal value is for k=0, e^(-π/2).

I just put this in the scientific calculator, and got the answer.

Try not to be so suspicious please. I genuinely love solving riddles.

Now the main question becomes i^i. The base i can be written as e^iθ. This makes it

e^[(i^2)θ]

= e^-θ

According to what we've been taught, the general value for θ is kπ + π/2. The principal value is for k=0, e^(-π/2).

I just put this in the scientific calculator, and got the answer.

Try not to be so suspicious please. I genuinely love solving riddles.

Reply by Caprico
- March 14 2011 01:50:16 AM

Sorry, I missed the last step. e^(-π/2) will give you the value of i^i. As we required [i^i]^2, just square it. Its equivalent to e^(-π).

Reply by Caprico
- March 14 2011 01:50:53 AM

The symbols π and θ courtesy of Wikipedia.

Reply by Ragib
- March 14 2011 06:17:00 AM

i m 17 too.hsc examinee 4m bangladesh.will try 4 buet.got any fb account?4 ur infrmtion,eulers formula is e^ix=cosx+isinx.if x=pi,then e^i.pi=-1=>(e^i.pi)^i=(-1)^i=>e^-pi=(-1)^i

Reply by Caprico
- March 14 2011 11:04:40 AM

You've juz used x instead of theta. the meaning is same. Notations dnt matter... And we've been taught in the way I showed you, nothing wrng wid dt!

Reply by Caprico
- March 14 2011 11:21:20 AM

In fact, if you want, just substitute the general formula I've written in place of x. You'll get values of cosx + isinx... Our notations are different, but our solutions are the same...

Reply by Ragib
- March 14 2011 01:38:43 PM

i didn't say that you are wrong.but i went a little easier.and,you said that it's riddle site.but it is full of probability problems and lots of other mathematical problem.i apologise for the latest problem i gave.try my football riddle.it's my genuine one.i created it.

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