## i^i

For college and university level advanced Mathematics
AntiviruShahriar
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### i^i

$e^{ix} = cos x + i \cdot sin x$ ব্যবহার করলে আমরা $i^i$ এর মান বের করতে পারি। এবং মান পাই $i^i=e^{\frac{- \pi}{2}}$
$i$ একটা অবাস্তব সংখ্যা। একটা অবাস্তব সংখ্যার অবাস্তব তম ঘাত কিভাবে বাস্তব হইতে পারে???

Moon
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### Re: i^i

হতে যে পারে সেইটা তো তুমি নিজেই প্রমাণ করলা! $e^{i\pi/2}=i \Longrightarrow i^i=e^{\frac{- \pi}{2}}$
"Inspiration is needed in geometry, just as much as in poetry." -- Aleksandr Pushkin

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AntiviruShahriar
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### Re: i^i

বুঝলাম না
$e^{i \frac{\pi}{2}}$ কি অবাস্তব না বাস্তব???এইটা নিয়া ও তো এখন সন্দেহ হইতাসে
সবই এখন সন্দেহজনক লাগতেসে। পড়তে যাই, আমার মাথা কেমন ভারি ভারি লাগতেসে।

Moon
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### Re: i^i

বেশি সন্দেহ করার কিছু নাই।

তবে আমারও একটা প্রশ্ন আছে বড়দের প্রতি। $a^b$ এর মানে কী? $b \in \mathbb{Z}$ হইলে আমরা বলতে পারি যে $b$ সংখ্যক $a$ কে গুণ করা। কিন্তু অন্যান্য ক্ষেত্রে formal definition টা কী?
"Inspiration is needed in geometry, just as much as in poetry." -- Aleksandr Pushkin

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tanvirab
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### Re: i^i

$i^i$ বাস্তব হইতে কোনো সমস্যা নাই। এবং বাস্তব যে সেইটাতো তুমি প্রমাণই করলা।

যেকোনো দুইটা জটিল সংখ্যা $a,b$ এর জন্য $a^b = e^{b \mbox{ } log(a)}$ যেইখানে exponential এর সংজ্ঞা দেয়া হইসে power series এর মাধ্যমে এবং logarithm এর সংজ্ঞা দেয়া হইসে exponential এর inverse হিসেবে।

nayel
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### Re: i^i

The definition that I know of $x^a$ is $x^a=e^{a\ln x}=\exp(a\ln x)$, where $\exp$ is defined as
$\exp(x)=\sum_{n=0}^\infty\frac{x^n}{n!}.$
Note that this definition generalizes to complex powers as well.

However, from your point of view: If $a$ is a positive integer, we regard $x^a$ as $\underbrace{x\cdot x\cdots x}_{a\text{ times}}$. From here it's clear that
$x^{m+n}=x^m\cdot x^n\qquad (*)$
for positive integers $m,n$. We can define negative powers from $(*)$, for example $x^{n-1}=x^{-1}\cdot x^n$ so $x^{-1}=1/x$. From $(*)$ we can also see that
$(x^m)^n=x^{mn}\qquad (**)$
for integers $m,n$. Now, from $(**)$ we can define $x^{1/m}$ as $y$, which is the solution to $y^m=(x^{1/m})^m=x$. Thus, for rational $a$, i.e. $a=p/q$ we have $x^a=(x^p)^{1/q}$. But this stops here, because we can't define irrational powers in this way.

P.S. Tanvir bhai beat me
"Everything should be made as simple as possible, but not simpler." - Albert Einstein

Moon
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### Re: i^i

ধন্যবাদ। ভাল একটা জিনিস শিখলাম।

তানভীর ভাই:
তবে এইখানে একটা পুরানো প্রশ্ন চলে আসছে। সেইটা হল জটিল সংখ্যার লগারিদম কেমন করে define করব? exponential এর inverse power series হিসেবে?
"Inspiration is needed in geometry, just as much as in poetry." -- Aleksandr Pushkin

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Moon
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### Re: i^i

oups...sorry...I think that we have posted in Bangla in Olympiad forum. BTW I think that this topic is suitable for college forum. It has nothing to with Olympiad problems/NT. moving it.
"Inspiration is needed in geometry, just as much as in poetry." -- Aleksandr Pushkin

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tanvirab
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### Re: i^i

Logarithm is defined as the inverse of the exponential function. So, $log(z) = w$ such that $e^w = z$.

But, one needs to be careful since the exponential of complex numbers is not a one-to-one function. It's a many-to-one function and the inverse is not automatically a well defined map. So, we need to make a choice. (This is similar to the inverse trigonometric functions, where we need to choose a principle value)

The usual convention is to note that, if we represent complex numbers in polar form, $z = re^{i\theta}$ and restrict $\theta$ to $[0, 2\pi)$ then the above definition is equivalent to $log(z) = log(r) + i \theta$. ($r$ is positive real, the function is not defined for $z=0$)

This function can also be expressed as a power series, and it is the obvious generalization of the Taylor series for the usual logarithm function. But, there is a technical complication in doing this, known as branch cuts;