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I want to find out the solutions to the equation $a^b = 1$. I know the real solutions like $k^0 = 1, 1^k = 1, (-1)^{2k} = 1$. I want to know if I missed any real solutions for this. Also I know of complex solutions like $1,\omega, \omega^2, ......$.

The actual question which I wanted to solve using this is $(x^2 - 5x + 6) ^ {x^2 - 3x + 2} = 1$. I have the solutions that can be obtained by the above real solutions to $a^b = 1$. I want any other possible real or complex solutions for the second equation.

PS: My main concern is about the second equation only.

Edit 1: I have tried to solve this by the general forms of complex numbers. Like in

$\left(re^{i\theta}\right)^{\left(a+ib\right)}\ =\ 1$. This simplifies to: $\left(\frac{r^a}{e^{\theta b}}\right)\left(e^{\theta a}.r^b\right)^i\ =\ 1$

$( r.e^{(i*\theta)} ) ^ {(a+ib)} = 1.\;$ This simplifies to: $((r^a)/e^{(\theta . b)}) . (e^{(\theta . a)}*r^b)^i = 1.$

Edit 2: Are there any purely complex numbers ( in the form $(m + i*n)$ where $n \neq 0$ ) such that $k^i\ =\ 1$

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    $\begingroup$ Note that for $k^0=1$ you need to make sure that $k\ne 0$ $\endgroup$ – ℋolo Jul 19 '18 at 16:16
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    $\begingroup$ What is your definition of $a^b$ ? In particular, what number spaces do $a$ and $b$ belong to ? $\endgroup$ – G. Fougeron Jul 19 '18 at 16:16
  • $\begingroup$ Hint: any complex number can be written as $z=Re^{i\theta},\mbox{where } R,\theta\in\Bbb R$ $\endgroup$ – ℋolo Jul 19 '18 at 16:20
  • $\begingroup$ @g-fougeron Both a and b belong to the set of complex numbers. $\endgroup$ – kaushalpranav Jul 19 '18 at 16:24
  • $\begingroup$ So $b$ is not a positive integer? $\endgroup$ – fleablood Jul 19 '18 at 16:39
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$$\mathbf{\color{green}{Full\ edition.}}$$

$$\mathbf{\color{brown}{The\ issue\ equation,\ real\ roots.}}$$

Let us consider the equation $$(y^2-y)^{y^2+y}=1,\quad \text{where } y=x-2.\tag1$$ The first special solution is $y_0=0,$ and the second is $y_1=-1.$ Then $$y^2-y=1,\quad y_{3,4}=\frac{1\pm\sqrt5}2,$$ and there are all real roots.

$$\mathbf{\color{brown}{Equation\ a^a=1,\ complex\ roots.}}$$ This is a test example. The real roots are $0$ and $1.$

Let $$a=re^{is},\tag2$$ then \begin{align} &a\log a = 2\pi mi,\\ &(r\cos s+ir\sin s)(\log r+is)=2\pi m i,\\ &\begin{cases} r\cos s\cdot\log r - r\sin s\cdot s=0\\ r\sin s\cdot \log r + r\cos s\cdot s = 2\pi m, \end{cases}\Rightarrow &\begin{cases} \log r = s\tan s\\ r s = 2\pi m \cos s, \end{cases}\Rightarrow &\begin{cases} r\log r = 2\pi m\sin s\\ r s = 2\pi m \cos s, \end{cases}\tag3 \end{align} with the numeric solutions $$\begin{pmatrix}m\\r\\s\end{pmatrix}= \left\{ \begin{pmatrix}1\\3.82059515\\0.95282627\end{pmatrix}, \begin{pmatrix}2\\6.04151895\\1.04462732\end{pmatrix}, \begin{pmatrix}3\\8.02197691\\1.08893606\end{pmatrix}, \begin{pmatrix}4\\9.86707599\\1.11688251\end{pmatrix},\cdots \right\}\tag4.$$

Exact solutions $e^{W(2\pi m i)}$ are the same.

$$\mathbf{\color{brown}{Equation\ (y^2-y)^{y^2+y}=1,\ complex\ roots.}}$$

The complex roots can be found from the equation $$(y^2+y)\mathrm{Ln}(y^2-y)=2\pi mi.\tag5$$ Let $$y=u+iv,\tag6$$ then \begin{align} &y^2\pm y = u^2-v^2\pm u + i(2uv\pm v),\\[4pt] &|y^2\pm y|^2 = |y|^2|y\pm1|^2 = (u^2+v^2)(u^2+v^2\pm 2u+1),\\[4pt] &\mathrm{Ln}(y^2-y)=\dfrac{2\pi mi}{|y^2+y|^2}(\Re(y^2+y)-i\Im(y^2+y)) = \dfrac{2\pi m}{|y^2+y|^2}(\Im(y^2+y)+i\Re(y^2+y))\\[4pt] & = \dfrac{2\pi m(2uv+v)}{(u^2+v^2)(u^2+v^2+2u+1)} + i\cdot \dfrac{2\pi m(u^2-v^2+u)}{(u^2+v^2)(u^2+v^2+2u+1)},\\[4pt] &u^2-v^2-u + i(2uv-v) = \exp\ \dfrac{2\pi m(2uv+v)}{(u^2+v^2)(u^2+v^2+2u+1)}\\[4pt] &\times\left(\cos\ \dfrac{2\pi m(u^2-v^2+u)}{(u^2+v^2)(u^2+v^2+2u+1)}+i\sin\dfrac{2\pi m(u^2-v^2+u)}{(u^2+v^2)(u^2+v^2+2u+1)}\right), \end{align} \begin{cases} (u^2+v^2)(u^2+v^2-2u+1) = \exp\ \dfrac{4\pi m(2uv+v)}{(u^2+v^2)(u^2+v^2+2u+1)}\\[4pt] u^2-v^2-u = \exp\ \dfrac{2\pi m(2uv+v)}{(u^2+v^2)(u^2+v^2+2u+1)} \cos\ \dfrac{2\pi m(u^2-v^2+u)}{(u^2+v^2)(u^2+v^2+2u+1)}\\[4pt] 2uv-v = \exp\ \dfrac{2\pi m(2uv+v)}{(u^2+v^2)(u^2+v^2+2u+1)} \sin\ \dfrac{2\pi m(u^2-v^2+u)}{(u^2+v^2)(u^2+v^2+2u+1)}\\[4pt] \end{cases} or, taking in account the identity $$\tan \dfrac t2=\dfrac{\sin t}{1+\cos t},\tag7$$ \begin{cases} (u^2+v^2)(u^2+v^2+2u+1)\log((u^2+v^2)(u^2+v^2-2u+1)) = 4\pi m(2uv+v)\\[4pt] \tan\ \dfrac{\pi m(u^2-v^2+u)}{(u^2+v^2)(u^2+v^2+2u+1)} = \dfrac{2uv-v}{\sqrt{(u^2+v^2)(u^2+v^2-2u+1)} + u^2-v^2-u}.\tag8\\ \end{cases} The system $(8)$ can be solved numerically.

$$\mathbf{\color{brown}{Equation\ (y^2-y)^{y^2+y}=1,\ solutions.}}$$

$\mathbf{\color{gray}{Case\ m=1.}}$

The system $(8)$ was solved, using MathCad, with the results $$y=\{-1.9945685-0.9447719i,\ -0.1678984+1.4557984i,\ 1.7735670+0.6139028i\}.$$ The plot of the first equation and of the second one are

m=1, First enter image description here

$\mathbf{\color{gray}{Case\ m=1.}}$

The system $(8)$ was solved, using MathCad, with the results $$y=\{-2.4059865-1.2679361i,\ 0.0090403+1.9177304i,\ 2.0491421+1.0079400i\}.$$ The plots of the first equation and of the second one are

m=2 First m=2 Second

Solutions were verified using Wolfram Alpha, with the positive results.

$$\mathbf{\color{brown}{How\ many\ solutions?}}$$

Let us show that for any $m$ there are at least one solution.

Really, let $$F(u,v) = (u^2+v^2)\log(u^2+v^2)+(u^2+v^2)\log\left(u^2+v^2-2u+1\right) - \dfrac{4\pi m(2uv+v)}{u^2+v^2+2u+1},$$ then $$F\left(\dfrac12,\dfrac12\right)=\dfrac12\log\dfrac12+\dfrac12\log\dfrac14 -\dfrac{4\pi m}{\dfrac52}=-\dfrac32\log2-\dfrac85\pi m <0,$$ $$(u^2+v^2)\log\left(u^2+v^2-2u+1\right)>-\log2,$$ $$\dfrac{2uv+v}{u^2+v^2+2u+1}<1$$ (see also Wolfram Alpha) $$(u^2+v^2)\log(u^2+v^2) -\log2-4\pi m \le F(u,v) \le 2(u^2+v^2)\log(u^2+v^2) -\dfrac85\pi m, $$ so there is a part of the the graph $F(u,w)=0$ inside of the area $$\left(e^{\frac12W(\frac45\pi m)}\right)^2 \le u^2+v^2 \le \left(e^{\frac12W(4\pi m+ \log2)}\right)^2,\quad u\ge\dfrac12,\quad v\ge\dfrac12,\tag9$$ where W(t) is Lambert W function (see also Wolfram Alpha plot).

Similarly, let us consider the function $$G(u,v)=\tan\ \dfrac{\pi m(u^2-v^2+u)}{(u^2+v^2)(u^2+v^2+2u+1)} - \dfrac{(2u-1)v}{\sqrt{(u^2+v^2)(u^2+v^2-2u+1)} + u^2-v^2-u}.$$ Using AM-GM for $u>\frac12$, one can get $$\sqrt{(u^2+v^2)(u^2+v^2-2u+1)}+u^2-v^2-u \le u^2+v^2-u+\dfrac12 = \dfrac14((2u-1)^2+1),$$ $$(u^2+v^2)(u^2+v^2+2u+1) \le (u^2+v^2+u+\frac12)^2 = \dfrac14((u+\frac12)^2+\frac14+v^2),$$ or, taking in account inequality $$\dfrac t{t^2+1}<\dfrac12,$$ $$G(u,v)\ge\tan\left(\pi ms(u,v)\right)- 2v,$$ where $$s(u,v)=\dfrac{(u^2-v^2+u)}{(u^2+v^2)(u^2+v^2+2u+1)}.$$ If $m>2,$ then the line $2\pi ms(u,v)=\dfrac \pi2$ is inside the area $(9)$ (see also Wolfram Alpha plot for $m=200.$)

The area of solution

Easily to see that the solution is in the calculated area.

This shows that for each $m$ there are at least one solution.

Therefore, $\mathbf{\color{green}{there\ are\ infinitely\ many\ complex\ solutions\ of\ the\ equation}\\\color{brown}{(x^2-5x+6)^{x^2-3x+2} = 1.}}$

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Let's find all of complex solutions. Let $a=re^{i\theta}$ and $b=x+iy$therefore $$a^b=r^xe^{-\theta y}r^{iy}e^{i\theta x}=1$$ which leads to $$r^x=e^{\theta y}\\r^{iy}e^{i\theta x}=1$$which by simplification gives us$$x\ln r=\theta y\\y\ln r+ \theta x=2k\pi\qquad,\qquad k\in\Bbb Z$$we are looking for the solutions where $\theta=0,\pi$ and $y=0$. Substituting $y=0$ gives us$$x\ln r=0\\\theta x=2k\pi$$

If $r\ne 1$ then $\ln r\ne 0$ and we have the trivial answer $x=0$ or $b=0$ and $a\ne 0$.

If $r=1$ and $\theta=0$ the 2nd equality is satisfied with $k=0$ and we have the trivial answer $a=1$ and any $b\in \Bbb R$.

If $r=1$ and $\theta=\pi$ we have $$x=2k\quad,\quad\forall k$$which gives us another trivial answer $a=-1$ and $b=2k$ where $k$ is integer, therefore

$$\\\LARGE\text{The trivial answers in real numbers are the only ones.}$$

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