I was using Euler's formula $ e^{ix} = \cos(x) + i \sin(x)$ with the case when $ x = 2\pi$. $$ e^{i2\pi} = \cos(2\pi) + i\sin(2\pi) = 1 + 0i $$ So I end up with the identity $ e^{i2\pi} = 1$. Now, if I apply $\ln$ it goes $$ \ln e^{i2\pi}=\ln1$$ $$ i2\pi = \ln1$$ How $\ln1=0$, Should be that $i2\pi=0$ its true, or not? The first implication I found its that you can divide any imaginary number by zero. $$ {ai\over 0} = {ai\over i2\pi} = {a\over 2\pi}$$ What don you think?
90400
I was trying to solve for $i^i$.  I got distracted and did the following: $e^{i \theta} = \cos(\theta) + i\sin(\theta)$ $e^{i 2\pi} = cos(2\pi) + i\sin(2\pi) = 1$ ||| Take ln() of two sides of equation: $\ln(e^{i2\pi}) = \ln(1)$ $i2\pi = 0$ Clearly this is false. What did I do wrong?
43632
I've been stuck with this for a while now. I have this chain of reasoning that would imply $e^{-\pi}=e^\pi$, obviously false, since $e^\pi$ and $e^{-\pi}$ are two real distinct numbers and so I must have made an assumption somewhere that I cannot actually do. I know I have to be very careful when working with complex numbers, especially when they're in the exponents, and so I tried to make the steps as small as I could so that it would be easier to point out where it went wrong.  \begin{align} e^{-\pi}&amp;= e^{\pi\cdot -1}\tag{1}\\ &amp;=e^{\pi\cdot i^2}\tag{2}\\ &amp;= e^{\pi\cdot i\cdot i}\tag{3}\\ &amp;= \left(e^{\pi\cdot i}\right)^i\tag{4}\\ &amp;= (-1)^i\tag{5}\\ &amp;=\left(\tfrac{1}{-1}\right)^i\tag{6}\\ &amp;=\left((-1)^{-1}\right)^i\tag{7}\\ &amp;=(-1)^{-1\cdot i}\tag{8}\\ &amp;=(-1)^{-i}\tag{9}\\ &amp;=\left(e^{i\pi}\right)^{-i}\tag{10}\\ &amp;=e^{i\pi\cdot -i}\tag{11}\\ &amp;=e^{-i^2\pi}\tag{12}\\ &amp;=e^\pi\tag{13} \end{align} I suspect it has something to do with changing the base from $e$ to $-1$, but what does that mean? Are complex powers only defined for positive bases? Any help is appreciated.
85874
Hi I was asked by a friends child who is in middle school why $e^{i\pi}+1=0$. Now I couldn't think of a way to explain it so he would understand. Albert Einstein once said “If you can't explain it simply, you don't understand it well enough” so how would I explain it.
79419
Euler's identity is $e^{i \pi} + 1 = 0$, a special case of the Euler's formula $e^{i\theta} = \cos\theta + i\sin\theta$, where $\theta = \pi$ (half-turn of the unit circle).  It is commonly described as "beautiful" because it is simple and includes a bunch of fundamental mathematical constants and operators, including the complex numbers.    Is there a corresponding Euler's identity for ?  Speculation:    proposes a different fundamental circle constant, $\tau = 2 \pi$, and a corresponding alternate Euler identity, based on a complete turn, the special case of $\theta = \tau$:   $e^{i\tau} = 1$  If both identities are extended to quaternions, is one any simpler or more "beautiful" than the other?
11814
Is it possible to prove that $\cos(\pi/3)=1/2$ without using any trigonometric identities?  To be specific, my calculus-book simply states that with an angle of $\pi/3$ on the unit-circle the points (0,0) and (1,0) are the vertices of an equilateral triangle with edge length 1. But how can one reach such conclusion?  Thanks!
105702
I know that $$\int^\infty_0\frac x{e^x-1}dx=\frac{\pi^2}{6}$$ For substituting $u=2$ into $$\zeta(u)\Gamma(u)=\int^\infty_0\frac{x^{u-1}}{e^x-1}dx$$  However, I suspect that there is an easier proof, maybe by the use of complex analysis. I haven't learnt the  yet. All I know is the above formula and $\zeta(2)=\frac{\pi^2}{6}$. Can we can use the above integral to find out some of the $\zeta$'s value?
35974
I found a solution but when testing it, it does not work....?    $$\cos z + \sin z = i \\ \frac{e^{iz} + e^{-iz}}{2} + \frac{e^{iz} - e^{-iz}}{2i} = i \\ e^{2iz}(1+i) + 2e^{iz} + i - 1 = 0 \\ e^{iz} = \frac{-1 \pm \sqrt{2-i}}{1+i} \\ z = -i \mathrm{Log}\left(\frac{-1\pm\sqrt{2-i}}{1+i}\right) + k2\pi$$    Upon testing it in W|A with $k=0$, I don't get the required result $(=i)$. I'm not sure where I went wrong and I did this a few times already.
34735
The problem reads as follows:     Noting that $t=\frac{\pi}{5}$ satisfies $3t=\pi-2t$, find the exact value of    $$\cos(36^\circ)$$   it says that you may find useful the following identities: $$\cos^2 t+\sin^2 t = 1,\\ \sin 2t = 2\sin t\cos t,\\ \sin 3t = 3\sin t - 4\sin^3 t. $$  Do I have to do a system of linear equations in function of ..what? $t$? $\cos$?  Thanks in advance :)
48474
I know that if $\sin x=\sin y$ then $x=n\pi+(-1)^ny$  and $\cos x=\cos y$ implies that $x=y+2k\pi$ or $x=-y+2l\pi$. Since the latter does not fit the form above $x=y+2k\pi$ which implies that $k=0,1,-1$. Hence, we have x=y or $x=2\pi,y=0$ or $x=0,y=2\pi$.
97081
This is where I'm at:  I know $$ \cos(z) = \frac{e^{iz} + e^{-iz}}{2} , \hspace{2mm} \sin(z) = \frac{e^{iz} - e^{-iz}}{2i}, $$ where $$ \tan(z) = \frac{\sin(z)}{\cos(z)}. $$ Applying the above, with a little manipulation, gives me: $$ \tan(z) = \frac{i\left(e^{-iz} - e^{iz}\right)}{e^{iz} + e^{-iz}}.$$  My thoughts are that I could use $e^{z} = e^{x+iy} = e^x\left(\cos(y) + i\sin(y) \right)$ to express both the numerator and denominator in trig form. Then I could times both by the denominator's complex conjugate as to get a real denominator, which would then, in turn, allow me to express the function in its real and imaginary parts.   However, as I'm sure you'd agree, this is messy, and it's hard to shake the idea that I'm missing something far more elegant.
48836
I'm currently self-studying &quot;Theory of Functions of a Complex Variable&quot; by A. I. Markushevich and now I've encountered several difficult problems in the book from the start. This is just one of them. Prove that every complex number $z$ where $|z|=1$ but $z\not=-1$ can be represented as $$z=\frac {1+it}{1-it}$$ where $t \in \mathbb{R}$. I've tried writing $$z=e^{i(t+\frac{\pi}{2}-\frac{\pi}{2})}$$ $$=\cos(t+\frac{\pi}{2}-\frac{\pi}{2})+i\sin(t+\frac{\pi}{2}-\frac{\pi}{2})$$ and then reducing by the addition formula but it doesn't seem to go anywhere. If someone would give me a hint to start, not the full answer, I would appreciate it very much.
19167
$$\zeta(it)=2it\pi it−1\sin(i\pi t/2)\Gamma(1−it)\zeta(1−it).$$ Everything on the RHS is never zero,  Does that means LHS has no zeros, since $\sin(s)$ has a simple zero at $s=0$ while $\zeta(1−s)$ has a simple pole at $s=0$  (the Laurent expansion), so the product $\sin(i\pi t/2)\zeta(1−it)$ is finite and nonzero at $t=0$.  My question is that from functional equation one can't get direct answer $\zeta(0)=−1/2$ unless using alternate infinite series !
8702
Find all roots in $\mathbb{C}$ of the equation $\cos(z)=2$ and plot them on an Argand diagram.  My solution: $\cos(z)=2$, $z=x+iy$ so we have that $\cos(x)\cos(iy)-\sin(x)\sin(iy)=2$ so $\cos(x)\cosh(y)=2$ and $\sin(x)\sinh(y)=0$ must hold. So if $\sinh(y)=0$ then $y=0$ so $\cosh(y)=1$ and so $\cos(x)=2$ which is a contradiction, therefore $x=n\pi$, for some $n\in \mathbb{Z}$ and so $\cos(x)=\pm 1$. However, $\cosh(y)&gt;0$, so if $\cos(x)=-1$ it follows that $\cos(x)\cosh(y)&lt;0$ and cannot equal $2$. Hence $\cos(x)=1$ so $x=2n\pi$ for some $n\in \mathbb{Z}$ and $y=\pm\cosh^{-1}(2)$ since $\cosh(y)=2$. Therefore since $z=x+iy$ $\implies z=2n\pi \pm i\cosh^{-1}(2)$, $n\in \mathbb{Z}$.  My question is, is this sufficent to answer the question $\cos(z)=2$? And if not how should I structure it using this method? Also why is it $\pm\cosh^{-1}(2)$? I don't understand this sutble point. And if I was to draw an Argand diagram would it just be on the imaginary axis at $\cosh^{-1}(2)$ every $2n\pi$?
8145
I have been self-studying CA and find it very interesting. So, working through problems in a book I have, I ran across  $$\int_{0}^{\infty}\frac{\sin(ax)}{e^{2\pi x}-1}dx=\frac{1}{4}\coth(a/2)-\frac{1}{2a}$$  and $$\int_{0}^{\infty}\frac{\sin(ax)}{e^{x}+1}dx=\frac{1}{2a}-\frac{\pi}{2\sinh(\pi a)}$$  For the former, I wrote it as $\frac{e^{aiz}}{e^{2\pi z}-1}$ and used a rectangular contour with vertices $0, \;\ R. \;\ R+i, \;\ i$  $e^{2\pi z}-1$ has poles at $ni$. Of these, I think $i$ only lies within the contour.   Unless I am in error, I then calculated the residue at $i$ to be $\frac{e^{-a}}{2\pi}$  So, $2\pi i(\frac{e^{-a}}{2\pi})=ie^{-a}$  Now, where I get hung up is setting up the integrals around the contour. Two of which should tend to 0 as $R\to \infty$.  Here is what I done.  $$\int_{0}^{R}\frac{e^{iax}}{e^{2\pi x}-1}dx+\int_{0}^{\infty}\frac{e^{ai(R+iy)}}{e^{2\pi (R+iy)}-1}idy+\int_{R}^{0}\frac{e^{ai(x+i)}}{e^{2\pi (x+i)}-1}dx+\int_{\infty}^{0}\frac{e^{ai(iy)}}{e^{2\pi iy}-1}dy=-\sinh(a)$$  I am unsure of the limits on the second and fourth integrals.  I am not so sure this is correct. The second and fourth ones, which represent the vertical sides,  should tend to 0 as $R\to\infty$. I hope :).   This than gave me $(1-e^{-a})\int\frac{e^{iax}}{e^{2\pi x}-1}dx=-\sinh(a)$.  Which does not look correct. I did manage to solve this using series and $\pi csc(\pi z)$, but the contour I am unsure of.   Can someone lend a hand here?. Any advice on either would be appreciated.   For the other one, I only posted it because it looks similar, but I believe is actually more involved and 'tougher' if you will. If I can get one, perhaps I can manage to evaluate the other.  With that one, I think the same rectangle with the same vertices can be used, but $\pi i$ would have to be avoided. Perhaps a Principal Value in there somewhere. But, I was told to use vertices $0, \;\ R, \;\ R+2\pi i, \;\ 2\pi i$  Since I am relatively new to CA, setting up those integrals is the confusing part.    I can do easier types of contour integrals, but want to learn more about these more challenging ones.   Thanks for any assistance.
2958
Why is $\sin(\alpha+\beta)=\sin\alpha\cos\beta+\sin\beta\cos\alpha$? How can we find the RHS if we don't know what it is? (instead of proving the identity itself) I could find a geometric solution in , but is there any solution that doesnt require drawing something? Edit: I now saw some nice proofs using $e$ and Euler's identity. I would appreciate anything NOT using them too, as a change. Out of curiosity, Is there a proof that says you actually "can't" prove the identity using only other simple 1-variable identities like $\sin^2\alpha+\cos^2\alpha=1$?
72973
I have to prove the following statement.     $$1+\cos{2\pi\over5}+\cos{4\pi\over5}+\cos{6\pi\over5}+\cos{8\pi\over5}=0$$   I have tried to use the sum of angles formula for cosine, but didn't get to a point where I'd be able to show that it is equal to $0$.
75595
I have been attempting to solve this integral for several days in preparation for an exam and keep reaching dead ends.  I(a,b) = $\int_0^{\pi/2} \ln(a^{2}\cos^2(x)+b^{2}\sin^2(x)) dx$, where $a,b&gt;0$  My best approach so far is defining:  $F(x) = I(x, b)$  Then, I know:  $F(b) = I(b, b) = \pi/2 * \ln(b^2)$  Then I am now attempting to find $F'(x)$ using differentiation under the integral sign, and integrating from b to a.  This is where I am stuck - no matter what I do, I cannot solve the integral. Am I in the right direction or should this be solved in a different manner? I also considered something elliptical due to the acos, bsin in the original integral, yet it has also reached a dead end.