I know Euler's formula and how to take complex exponents, but in it it's $e$ to an imaginary angle, not a number, it seems. From my understanding pi itself is not an angle, but $\pi$ radians is. And since cosine can only take in an angle, or at least a representation of one, and the variable is the same everywhere in Euler's formula, the exponent should be an angle and not a number. The question is then raised could I then say $e^{180i}=-1$? Surely this has a single numerical value though, right?   I've been bothered by this for a long time. Am I wrong or how can I take an exponent with a complex number?   Edit for the duplicate: I understand how to take them and how they're supposed work and be justifyed/proved, but raising something to the power of an angle which is what I thought was meant to be happening made me question how they actually worked that could allow this.
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How does one find the algebraic solution of a Complex number raised to the power of another Complex number?  Here is the work I have done so far, if there are any mistakes please inform me.  A real number with a Complex Exponent: $$A \in \Bbb R, \space z \in \Bbb C,\space z = x + iy$$ $$A^z = A^{x+iy} = A^xA^{iy}$$ $$\implies e^{iy\ln A} = A^{iy}$$ $$\exp(iy\ln A) = \cos (y\ln A)+i\sin(y\ln A)$$ $$A^z = A^x[\cos (y\ln A)+i\sin(y\ln A)]$$  A Complex number with a Complex Exponent: [Using previous variables] $$C \in \Bbb C,\space C = a +bi\space|\space re^{i\theta}, \theta =\arg C$$ $$C^z = (a+ib)^z$$ [After previous mistake the following notes are inaccurate] $$C^z = a^x[\cos (y\ln a)+i\sin(y\ln a)] + ib^xib^{iy}$$  And this is as far as I could go as the expansion of $ib^xib^{iy}$ is being problematic. Could I have some help with this expansion?
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We know Euler formula:  $$ e^{ix}=\cos(x)+i\sin(x). $$  Does this formula work if we replace real number $x$ with complex number $z$?
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I've a few questions in mind:   Why is $\pi$ irrational? If it is, then how can 2 rational quantities (circumference, diameter) can produce irrational number? How are we able to determine digits of $\pi$ accurately?
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Express $-1+i$ in exponential form.  My attempt so far  Let $z=-1+i$  $$r=|z|=\sqrt2$$  $$\theta=\tan^{-1}(-1)=-\frac{\pi}{4}$$  Now, this is where I go wrong (I don't know why it's wrong!):  So in exponential form: $-1+i=\sqrt2 e^{-i\pi/4}$  According to the solutions, $\theta=3\pi/4$.  Thanks in advance
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We have $$e^{2\pi i n}=1$$  So we have $$e^{2\pi in+1}=e$$  which implies $$(e^{2\pi in+1})^{2\pi in+1}=e^{2\pi in+1}=e$$ Thus we have $$e^{-4\pi^{2}n^{2}+4\pi in+1}=e$$  This implies $$e^{-4\pi^{2}n^{2}}=1$$  Taking the limit when $n\rightarrow \infty$ gives $0=1$.
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From Euler we've learned that $z=re^{i\theta}$.  And it's easy to see that $|z|^2=r^2$, since $re^{i\theta}\times re^{-i\theta}=r^2$.  Why must we use e to represent these numbers correctly? It seems that I could arbitrarily choose a different exponent $z=r\pi^{i\theta}$ and get the same size for $z$ as I did before: $|z|^2=r\pi^{i\theta}\times r\pi^{-i\theta}=r^2$  What did I miss?
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What's wrong with this : $$e^{i\pi} = -1$$ $$\therefore e^{2i\pi} = 1$$ $$\therefore \log \left( e^{2i\pi} \right ) = \log(1) = 0$$ $$\therefore 2i\pi = 0$$
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I'm having trouble figuring out the Fourier series of $|\cos(x)|$ from $-\pi$ to $\pi$.  I understand its an even function, so all the $b_n$s are $0$  $$a_0 = \frac 2 \pi \int_0^\pi |\cos(x)|\,dx = 0$$  $$a_n = \frac 2 \pi \int _0^\pi  |\cos(x)| \cos(nx) \, dx = \frac 2 \pi \int_0^\pi \cos^2(x)\,dx.$$  since for all $j,k$ not equal the integral is zero.  so only $a_1$ remains. is this correct?   How would I evaluate $\sum_{n=1}^\infty (-1)^{n-1} /(4n^2 - 1)\  {}$?
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We know that $e^{\pi i} = -1$ because of de Moivre's formula. ($e^{\pi i} = \cos \pi + i\sin \pi = -1).$  Suppose we square both sides and get $e^{2\pi i} = 1$(which you also get from de Moivre's formula), then shouldn't $2\pi i=0$? What am I missing here?
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Use the polar form of complex numbers to show that every complex number $z\neq0$ has multiplicative inverse $z^{-1}$.  If $z=a+bi$, then the polar form is $z=r(cos(\alpha))+i(sin(\alpha))$.  I can do it, not using polar coordinates: Let $z=a+ib$.  Since $1+0i$ is the multiplicative identity, if $x+iy$ is the multiplicative inverse of $z=a+ib$, then $(a+ib)(x+iy)=1+i0$ $\Rightarrow$ $(ax-by)+i(ay+bx)=1+i0$ $\Rightarrow$ $ax-by=1bx+ay=0$ $\Rightarrow$ $x=\frac{a}{a^2+b^2}$, $y=\frac{-b}{a^2+b^2}$, if $a^2+b^2\neq0$.  The multiplicative inverse of $a+ib$ is therefore $\frac{a}{a^2+b^2} +i\frac{-b}{a^2+b^2}$.  Now, should I use this same process and just replace $z=a+ib$ with $z=r(cos(\alpha))+i(sin(\alpha))$??  Or is there another way to go about this process?
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Since $i = \sqrt{-1}, i^2 = -1, i^3=-i, i^4 =1$ I understand to calculate, say,  $i^{999}$ I just have to $i^{999} = i^{4 \cdot 249 + 3} = (i^4)^{249} \cdot i^{3} = -i$ But I have a question here, why can't I do something like $i^{999} = (i^4)^{\frac{999}{4}} = 1$ clearly the answer is wrong, infact everytime I'll only end up with $1$. Can anyone point out which step exactly is invalid and with reasons like why is it wrong.   Which naturally extends the question to, how do you calculate $i$ to the power of any "rational" exponent. Like $i^{743/5}$ or something.  Thanks.
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Possible Duplicate:          I need to know why Euler's formula is true? I mean why is the following true: $$ e^{ix}  = \cos(x) + i\sin(x) $$
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I understand that when you raise any number $x$ to a power, you multiply $x$ by itself the number of times indicated in the power. However, what happens when $i^i$ is performed? How can a number be multiplied an imaginary amount of times? Wolfram Alpha says that it is equal to $e^{{-\pi}/{2}}$, but how would you arrive at that answer? Any response will be appreciated, thanks!
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Since $e^{ix}=\cos x+i\sin x$, thus  $e^{2\pi i}=\cos2\pi+i\sin2\pi=1$     Now I take arbitrary real number $r$ then $e^{i2\pi r}=(e^{i2\pi})^r=1^r=1?$    But this cannot be true since $\cos2\pi r+i\sin2\pi r$  isn't necessarily $1$ for arbitrary $r$. Where did it go wrong?
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I'm still struggling to understand why the derivative of sine only works for radians. I had always thought that radians and degrees were both arbitrary units of measurement, and just now I'm discovering that I've been wrong all along! I'm guessing that when you differentiate sine, the step that only works for radians is when you replace $\sin(dx)$ with just $dx$, because as $dx$ approaches $0$ then $\sin(dx)$ equals $dx$ because $\sin(\theta)$ equals $\theta$. But isn't the same true for degrees? As $dx$ approaches $\theta$ degrees then $\sin(dx \,\text{degrees})$ still approaches $0$. But I've come to the understanding that $\sin(dx \,\text{degrees})$ approaches $0$ almost $60$ times slower, so if $\sin(dx \,\text{radians})$ can be replaced with $dx$ then $\sin(dx \,\text{degrees})$ would have to be replaced with $(\pi/180)$ times $dx$ degrees.  But the question remains of why it works perfectly for radians. How do we know that we can replace $\sin(dx)$ with just $dx$ without any kind of conversion applied like we need for degrees? It's not good enough to just say that we can see that $\sin(dx)$ approaches $dx$ as $dx$ gets very small.  Mathematically we can see that $\sin(.00001)$ is pretty darn close to $0.00001$ when we're using radians. But let's say we had a unit of measurement "sixths" where there are $6$ of them in a full circle, pretty close to radians. It would also look like $\sin(dx \,\text{sixths})$ approaches $dx$ when it gets very small, but we know we'd have to replace $\sin(dx \,\text{sixths})$  with $(\pi/3) \,dx$ sixths when differentiating. So how do we know that radians work out so magically, and why do they?  I've read the answers to  and followed the links, and no, they don't answer my question.
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Should I write complex numbers as $a+ib$ or $a+bi$? Maybe there are certain cases where I should write it one way and other cases where I should write it the other way. What do you think?
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Why does $|e^{ix}|^2 = 1$?   The book said $e^{ix} = \cos x + i\sin x$, and square it, then $|e^{ix}|^2 = \cos^2x + \sin^2x = 1$.  But, when I calculated it, $ |e^{ix}|^2 = \left|\cos x + i\sin x\right|^2 = \cos^2x - \sin^2x + 2i\sin x\cos x$.  I can't make it to be equal $1.$ How can I do it?