| | | | | |  | | Ed Bell, former Senior Professional II, Server Engineer at Concentrix (1995-2020) |
| | There's no objective way in which OCaml is better than Haskell, but I can tell you why I like it more. In short, Haskell is too perfect. OCaml is better because of its imper... |
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First of all, let us try to express [math]i=\sqrt{\,-\,1}[/math] as a square of some number. That will make things easier for us, as we will soon see. When [math]i=\sqrt{\,-\,1}[/math] , it also means that, [math]{{i}^{2}}=\,\,-\,1[/math] . Then, observe that, [math]i=\frac{1}{2}\times \left( 2\times i \right)=\frac{1}{2}\times \left( 1+2\times i-1 \right)[/math] |
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Given that, [math]i=\sqrt{\,-\,1}\,\,,[/math] so that [math]{{i}^{\,2}}=\,\,-\,1[/math] and [math]{{i}^{\,3}}=\,\,-\,\,i\,\,.[/math] Now suppose that, [math]{{i}^{\,\frac{1}{6}}}=z\,\,,[/math] so that [math]{{z}^{6}}=i\,\,,[/math] that is, [math]{{z}^{6}}-i=0\,\,,[/math] that is, [math]{{z}^{6}}+{{i}^{3}}=0\,\,,[/math] that is, [math]{{\left( {{z}^{\,2}} \right)}^{3}}+{{i}^{3}}=0\,\,\Rightarrow \,\left( {{z}^{\,2}}+i \right)\times \left( {{\left( {{z}^{\,2}} \right)}^{2}}-{{z}^{\,2}}\times i+{{i}^{\,2}} \right)=0[/math] , using factorization formula. [math]\Rightarrow \,\,\left( {{z}^{\,2}}+i \right)\times \left( {{z}^{\,4}}-{{z}^{\,2}}\times i-1 \right)=0[/math] [math]\Rightarrow \,\,{{z}^{\,2}}+i=0\,\,,\,\,{{z}^{\,4}}-{{z}^{\,...[/math] |
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