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I was working on some basic complex numbers and suddenly have a question about it. As you all know, the complex numbers are created as we cannot find a place in the real number system to fit in the value (-1)^(1/2).

So we use i to represent it, but how do we know that any number to the power of any other number (including fraction, negative and others) are able to be expressed in a+bi? Do we need another number system for like (-1)^(1/4)? Although that is not a very good example, because we are able to show that (1/2 + i/2)^4 = -1 after some working.

After doing some researches on this topic, I think I come to the conclusion that we don't need another complex system but I don't know how to prove it. Can anyone tell me how?

I really appreciate any help on this.

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    $\begingroup$ Hint: By the en.wikipedia.org/wiki/Fundamental_theorem_of_algebra the complex numbers are algebraically closed. $\endgroup$ – gammatester Dec 5 '13 at 11:17
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    $\begingroup$ To elaborate on @gammatester's hint: Whatever we want to introduce by the name $(-1)^{1/4}$, it should be a root of the polynomial $X^4+1$, and that poolynomialk does have roots already in $\mathbb C$. $\endgroup$ – Hagen von Eitzen Dec 5 '13 at 11:22
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any complex number can be written in the form $$a+ib\to r(\cos\theta+i\sin\theta)=re^{i\theta}$$ if you raise the term to $n$ being any real or complex number. $$(a+ib)^n=r^ne^{in\theta}$$which is once more a complex no.

as per your need i will eloborate for a complex $n=c+id$ $$r^ne^{in\theta}=r^{c+id}e^{i(c+id)\theta}$$ $$ =r^c.r^{id}e^{-d\theta}e^{ic\theta}....(1)$$ now let $$r^{id}=e^{i\alpha}$$ $$i\alpha=id\ln r$$ therefore $$ r^{id}=e^{id\ln r}$$ substituting in $(1)$ $$(r^c.e^{-d\theta})e^{i(d\ln r+c\theta)}=(a+ib)^{c+id}$$ which is $$(r^c.e^{-d\theta})(\cos ( d\ln r+c\theta )+i\sin ( d\ln r+c\theta ))$$ which is finally of the form$$r'(a'+ib')$$

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  • $\begingroup$ Thanks for the answer. But there is still a bit where I don't understand, do you mean that (r^n)(e^inΘ) can still be express in a+bi even when n is a complex number. If that is the case, can you show me how? As I substitute n=(r)(e^iΘ) into the (r^n)(e^inΘ), I get a bunch of messy stuff. And I couldn't further simplify it into a+bi. $\endgroup$ – Theoretical Dec 5 '13 at 14:18
  • $\begingroup$ @Theoretical : i have given proof for a complex $n$. $\endgroup$ – Suraj M S Dec 5 '13 at 19:49

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