Consider Euler's Formula: $$e^{i\theta}=cos(\theta)+isin(\theta)$$ Using this, the value of $i$ is  $e^{{i\pi}/2}$. Following this: $$i^i=e^{i^2\pi/2}$$ The accepted principal result of this is that $$i^i=e^{-\pi/2}$$ I appreciate that the result given is one of many solutions - but for my question here, I'm only focussing on this case.  Instead of simply using $i^2=-1$, I will separate the exponent for $i^i=e^{i*i\pi/2}$.  Going back to Euler's Formula, and inserting ${\theta=i\pi/2}$ we get: $$i^i=\cos (i\pi/2) + i\sin (i\pi/2)$$ Which gives $$cosh(\pi/2)+i*sinh(pi/2)$$ Or roughly $2.509+2.301i$ Again - I know there are many different solutions here, I'm just focussing on the principal case.  Why is there a contradiction here? What exactly causes my method to not yield the same result? Originally my answer was marked as duplicate - however the thread I was linked to did not contain this method - and the reasons for its failure was said to be due to $ln(i)$ not being well defined over complex space (I used $ln(i)$ in my original question) . Considering that I can now show my method without the use of $ln(i)$, I am wondering if there is another reason why my method is incorrect.
2270
According to WolframAlpha,  but I don't know how I can prove it.
90759
Prove that $$\cos(\pi/11)+\cos(3\pi/11)+\cos(5\pi/11)+\cos(7\pi/11)+\cos(9\pi/11)=1/2$$ using Euler's formula.  Everything I tried has failed so far.  Here is one thing I tried, but obviously didn't work. $$\Re e \{e^{\frac{\pi}{11}i}(1+e^{\frac{2\pi}{11}i}+e^{\frac{4\pi}{11}i}+e^{\frac{6\pi}{11}i}+e^{\frac{8\pi}{11}i}) \}=\frac{1}{2}$$ $$\Re e \{e^{\frac{\pi}{11}i}(1+\sqrt[11]{e^{2\pi i}}+\sqrt[11]{e^{4\pi i}}+\sqrt[11]{e^{6\pi i}}+\sqrt[11]{e^{8\pi i}}) \}=\frac{1}{2}$$ $$\Re e \{e^{\frac{\pi}{11}i}(1+\sqrt[11]{1}+\sqrt[11]{1}+\sqrt[11]{1}+\sqrt[11]{1}) \}=\frac{1}{2}$$ $$\Re e \{5e^{\frac{\pi}{11}i} \}=\frac{1}{2}$$ $$5\cos(\frac{\pi}{11})=\frac{1}{2}$$ Which isn't true :D  Thanks in advance
25721
$$\sum_2^\infty e^{3-2n}$$  The formulas for these things are so ambiguous I really have no clue on how to use them.  $$\frac {cr^M}{1-r}$$  $$\frac {1e^2}{1-\frac{1}{e}}$$  Is that a wrong application of the formula and why?
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.
97512
To evaluate $e^{\sin(x)}$ I use the standard series $e^t$ and $\sin(t)$, combining them gives me:  $e^t = 1+t+\dfrac{t^2}{2!}+\dfrac{t^3}{3!}+\dfrac{t^4}{4!}+O(t^5)$  $\sin(t) = t-\dfrac{t^3}{3!}+O(t^5)$  $\therefore e^{\sin(x)} = 1+\sin(x)+\dfrac{\sin^2(x)}{2!}+\cdots+O(x^5)$  $\iff e^{\sin(x)}=1+\left(x-\dfrac{x^3}{6}\right)+\dfrac{\left(x-\frac{x^3}{6}\right)^2}{2}+\dfrac{x^3}{6}+O(x^5) = 1+x +\dfrac{x^2}{2}-\dfrac{x^4}{6} + O(x^5)$  In the last step, I only evaluate up to $\sin^3x$ term, everything above has a grade equal to or greater than $x^5$  However, I'm wrong. According to Wolfram, the series expansion for $e^{\sin x} = 1+x+\dfrac{x^2}{2}-\dfrac{x^4}{8}+O(x^5)$.  What did I do wrong? I can't seem to find what or where.
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 $2n\pi$?
19734
I was playing around with Euler's identity   $$\begin{align}e^{i\pi}&amp;=-1\\\\\frac{\log{(-1)}}{i}&amp;=\pi\\\\\frac{x \log(-1)}{i}&amp;=\pi x \end{align}$$   And when I had $x=2$  $$\frac{\log(1)}{i}=2\pi$$Which is equivalent to $$\frac{0}{i}=2\pi$$  How can this be?
92794
For a complex number $A$ and a real number $B$, when does the well-known formula $(e^A)^B = e^{AB}$ fail? Or does it hold at all for complex A?     Since $e^{2\pi i} = 1$, if this formula holds for all complex numbers $A$ and real numbers $B$, then it would imply that $e^{2\pi ti}=( e^{2\pi i} )^t = 1^t=1$ for all $t$, which is obviously false.
4664
For $a\ge 0$ let's define $$I(a)=\int_{0}^{\pi}\ln\left(1-2a\cos x+a^2\right)dx.$$ Find explicit formula for $I(a)$.   My attempt: Let  $$\begin{align*} f_n(x) &amp;= \frac{\ln\left(1-2 \left(a+\frac{1}{n}\right)\cos x+\left(a+\frac{1}{n}\right)^2\right)-\ln\left(1-2a\cos x+a^2\right)}{\frac{1}{n}}\\ &amp;=\frac{\ln\left(\displaystyle\frac{1-2 \left(a+\frac{1}{n}\right)\cos x+\left(a+\frac{1}{n}\right)^2}{1-2a\cos x+a^2}\right)}{\frac{1}{n}}\\ &amp;=\frac{\ln\left(1+\dfrac{1}{n}\left(\displaystyle\frac{2a-2\cos x+\frac{1}{n}}{1-2a\cos x+a^2}\right)\right)}{\frac{1}{n}}. \end{align*}$$ Now it is easy to see that $f_n(x) \to \frac{2a-2\cos x}{1-2a\cos x+a^2}$ as $n \to \infty$. $|f_n(x)|\le \frac{2a+2}{(1-a)^2}$ RHS is integrable so $\lim_{n\to\infty}\int_0^\pi f_n(x)dx = \int_0^\pi \frac{2a-2\cos x}{1-2a\cos x+a^2} dx=I'(a)$. But  $$\int_0^\pi \frac{2a-2\cos x}{1-2a\cos x+a^2}=\int_0^\pi\left(1-\frac{(1-a)^2}{1-2a\cos x+a^2}\right)dx.$$ Consider $$\int_0^\pi\frac{dx}{1-2a\cos x+a^2}=\int_0^\infty\frac{\frac{dy}{1+t^2}}{1-2a\frac{1-t^2}{1+t^2}+a^2}=\int_0^\infty\frac{dt}{1+t^2-2a(1-t^2)+a^2(1+t^2)}=\int_0^\infty\frac{dt}{(1-a)^2+\left((1+a)t\right)^2}\stackrel{(*)}{=}\frac{1}{(1-a)^2}\int_0^\infty\frac{dt}{1+\left(\frac{1+a}{1-a}t\right)^2}=\frac{1}{(1-a)(1+a)}\int_0^\infty\frac{du}{1+u^2}=\frac{1}{(1-a)(1+a)}\frac{\pi}{2}.$$ So $$I'(a)=\frac{\pi}{2}\left(2-\frac{1-a}{1+a}\right)\Rightarrow I(a)=\frac{\pi}{2}\left(3a-2\ln\left(a+1\right)\right).$$  It looks too easy, is there any crucial lack?  $(*)$ — we have to check $a=1$ here by hand and actually consider $[0,1), (1,\infty)$ but result on these two intervals may differ only by constant - it may be important but in my opinion not crucial for this proof.
84718
I am trying to establish that  $$\frac2{\pi}\int_0^{2\pi} f(e^{i\theta})\cos^2\left(\frac{\theta}{2}\right)\;\mathrm d\theta=2f(0)+f^\prime(0)$$  for $f$ analytic on $|z| &lt; R,\; R &gt; 1$  I think I am making a mistake somewhere, please point it out to me.Here's what I've done unsing the mean value property and Cauchy's integral formula:  $$2f(0)+f&#39;(0)=2(\frac1{2\pi}\int_0^{2\pi} f(e^{i\theta})d\theta + \frac1{2\pi i}\int_0^{2\pi}\frac{f(e^{i\theta})ie^{i\theta}}{e^{i2\theta}}=\frac{1}{\pi}\int_0^{2\pi} f(e^{i\theta})d\theta\frac1{2\pi}\int_0^{2\pi}\frac{f(e^{i\theta})}{e^{i\theta}}\quad\text{(1)}$$  But when I develop the expression $\frac2{\pi}\int_0^{2\pi} f(e^{i\theta})\cos^2(\frac{\theta}{2}) d\theta$ I get:  $$\frac1{\pi}\int_0^{2\pi} f(e^{i\theta})d\theta+\frac1{2\pi}\int_0^{2\pi}f(e^{i\theta})[{e^{i\theta}}+e^{i-\theta}]d\theta \quad\text{(2)}$$  The second integral in (1) differs from the second integral in (2). Thanks.
89218
I typed $\pi$ into Wolfram Alpha and in the short list of definitions there appeared  $$ \pi = -i \log(-1)$$  which really bothered me. Multiplying on both sides by $2i$:  $$ 2\pi i = 2 \log(-1) = \log(-1)^2 = \log 1= 0$$  which is clearly false.  I guess my error is $\log 1 = 0$ when $\log$ is complex-valued.  I need to use $1 = e^{2 \pi i}$ instead.  So my question: is it correct for WA to say $\pi = -i \log(-1)$?  Or should be they specifying "which" $-1$ they mean?  Clearly $ -1 = e^{i \pi}$ is the "correct" value of $-1$ here.
11328
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
22053
The formula we need to prove is $$ \displaystyle \iint_{x^{2}+y^{2} \leq 1} e^{x^{2}+y^{2}} d x d y \leq\left[\int_{-\frac{\sqrt{\pi}}{2}}^{\frac{\sqrt{\pi}}{2}} e^{x^{2}} d x\right]^{2}  $$ As I already done the left side $$ \iint_{x^{2}+y^{2} \leq 1} e^{x^{2}+y^{2}} d x d y  = \int_{-\pi}^{\pi} \mathrm{d}\varphi \int_{0}^{1} e^{r^2}r\mathrm{d}r  = 2\pi \int_{0}^{1} e^{r^2}r\mathrm{d}r = \left.\pi e^{r^2}\right|_{0}^{1} = \pi(e-1) $$ I still have no idea about the right side.
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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 $0=1$.
73212
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?
17860
I've already see a  by Marko Riedel which I list it follows:     The standard way to treat these sums is to integrate   $$ f(z) = \frac{1}{(z+\alpha)^2} \pi \cot(\pi z)$$   along a contour consisting of a circle of radius $R$ and with $R$ going to infinity and hence being larger than $\alpha$, where the circle does not pass through the poles on the real axis.   Now along the semicircle in the upper half plane we have   $$|f(z)| \le \frac{1}{(R-|\alpha|)^2}\pi \left|\frac{e^{i\pi R\exp(i\theta)} + e^{-i\pi R\exp(i\theta)}} {e^{i\pi R\exp(i\theta)} - e^{-i\pi R\exp(i\theta)}}\right|= \frac{1}{(R-|\alpha|)^2} \pi  \left|\frac{e^{2i\pi R\exp(i\theta)}+1}{e^{2i\pi R\exp(i\theta)}-1}\right| &lt; \frac{1}{(R-|\alpha|)^2} \pi   \frac{1+e^{-2\pi R\sin(\theta)}e^{2i\pi R\cos(\theta)}} {1-e^{-2\pi R\sin(\theta)}e^{2i\pi R\cos(\theta)}}$$   This last term is clearly $O(1/R^2)$ as $R$ goes to infinity as the quotient of the two exponentials goes to one since $\exp(-R)$ vanishes and there is no singularity when $\theta = 0$ or $\theta = \pi$ as $R\cos\theta = \pm R$, which is not an integer by the assumption that the circle avoids the poles and hence cannot be one.     My question: It is clear that for each $z=Re^{i \theta}$ , $f(z)=O(1/R^2)$ ，but can this guarantee that the last term on RHS is uniform bounded ? Since $\theta$ varies on a compact interval $[0, \pi]$ , I want to show that for each $ \theta \in [0, \pi]$ , and a fixed positive number $M \gt 1 $ there exist an open ball contains $\theta$ and a fixed $R_{\theta}\gt 0$ , for every $R \ge R_{\theta}$ and $x \in $ the open ball $$\frac{1+e^{-2\pi R\sin(x)}e^{2i\pi R\cos(x)}} {1-e^{-2\pi R\sin(x)}e^{2i\pi R\cos(x)}} \le M$$  For $\theta \neq 0,\pi$ , it is easy to find the desired $R_{\theta}$ and the open ball , but if $\theta = 0$ or $ \theta = \pi$ , for every open ball containing $\theta$ , it behave erratically near $\theta$ , I have no idea how to deal with this .
23507
I just came across the problem of what happens in the boundary cases of .  More concisely, for what $\theta$ will the following sum converge? $$\sum_{n=1}^{\infty} \frac{e^{ni\theta}}{n}$$ It is well known that this sum will diverge for $\theta=0$ and will converge for $\theta=\pi$. In fact, for any odd multiple of an even root of unity, since we have symmetry, for each line through the origin of the complex plane, we can use the alternating series theorem to show that they converge.  Another idea I had was to split  the sum into imaginary and real parts, and if we can prove each of these converges, then we are all set. So now we must prove that both of the following converge $$\sum_{n=1}^{\infty}\frac{\cos(n\theta)}{n}\space, \space \sum_{n=1}^{\infty}\frac{\sin(n\theta)}{n}$$ Now these can be shown to converge easily by Dirichlet's test for any rational multiple of pi, but what about irrational multiples of pi? I know it has been shown to converge with $\theta=1$, but I don't see how to generalize this.