Change Details

this equation array \begin{eqnarray} |z-z_i| <\delta_i/2,\\ |w-z_i| &=& |w-z+z-z_i| \nonumber\\ &\leq& |w-z| + |z-z_i| \nonumber \\ & <& \delta + \delta_i/2 \nonumber\\ & \leq& \delta_i/2 + \delta_i/2 \nonumber \\ &<& \delta_i. \end{eqnarray} is transofmred to \begin{align}I_n &=& \left|\int_{\partial \Delta_n} f(z)\,\mbox{d} z\right| \\ &= & \left|\int_{\partial \Delta_n} \left(f(z) - f(z_0) - (z-z_0)f'(z_0)\right)\,\mbox{d} z\right| \\ &\leqslant & \len(\partial \Delta_n)\varepsilon\sup_{z \in \partial \Delta_n}|z-z_0|.\end{align} but this is not the way of writing an align, since this would make the equations to be aligned on the right side of the page. The correct way of writing an align is: \begin{align}I_n &= \left|\int_{\partial \Delta_n} f(z)\,\mbox{d} z\right| \\ &= \left|\int_{\partial \Delta_n} \left(f(z) - f(z_0) - (z-z_0)f'(z_0)\right)\,\mbox{d} z\right| \\ &\leqslant \operatorname{length}(\partial \Delta_n)\varepsilon\sup_{z \in \partial \Delta_n}|z-z_0|.\end{align} The solution is not that easy, it is not trivial to chose which & remove for having the correct the alignment

this equation array ``` \begin{eqnarray} |z-z_i| <\delta_i/2,\\ |w-z_i| &=& |w-z+z-z_i| \nonumber\\ &\leq& |w-z| + |z-z_i| \nonumber \\ & <& \delta + \delta_i/2 \nonumber\\ & \leq& \delta_i/2 + \delta_i/2 \nonumber \\ &<& \delta_i. \end{eqnarray} ``` is transofmred to ``` \begin{align}I_n &=& \left|\int_{\partial \Delta_n} f(z)\,\mbox{d} z\right| \\ &= & \left|\int_{\partial \Delta_n} \left(f(z) - f(z_0) - (z-z_0)f'(z_0)\right)\,\mbox{d} z\right| \\ &\leqslant & \len(\partial \Delta_n)\varepsilon\sup_{z \in \partial \Delta_n}|z-z_0|.\end{align} ``` but this is not the way of writing an align, since this would make the equations to be aligned on the right side of the page. The correct way of writing an align is: ``` \begin{align}I_n &= \left|\int_{\partial \Delta_n} f(z)\,\mbox{d} z\right| \\ &= \left|\int_{\partial \Delta_n} \left(f(z) - f(z_0) - (z-z_0)f'(z_0)\right)\,\mbox{d} z\right| \\ &\leqslant \operatorname{length}(\partial \Delta_n)\varepsilon\sup_{z \in \partial \Delta_n}|z-z_0|.\end{align} ``` The solution is not that easy, it is not trivial to chose which & remove for having the correct the alignment

this equation array ``` \begin{eqnarray} |z-z_i| <\delta_i/2,\\ |w-z_i| &=& |w-z+z-z_i| \nonumber\\ &\leq& |w-z| + |z-z_i| \nonumber \\ & <& \delta + \delta_i/2 \nonumber\\ & \leq& \delta_i/2 + \delta_i/2 \nonumber \\ &<& \delta_i. \end{eqnarray} ``` is transofmred to ``` \begin{align}I_n &=& \left|\int_{\partial \Delta_n} f(z)\,\mbox{d} z\right| \\ &= & \left|\int_{\partial \Delta_n} \left(f(z) - f(z_0) - (z-z_0)f'(z_0)\right)\,\mbox{d} z\right| \\ &\leqslant & \len(\partial \Delta_n)\varepsilon\sup_{z \in \partial \Delta_n}|z-z_0|.\end{align} ``` but this is not the way of writing an align, since this would make the equations to be aligned on the right side of the page. The correct way of writing an align is: ``` \begin{align}I_n &= \left|\int_{\partial \Delta_n} f(z)\,\mbox{d} z\right| \\ &= \left|\int_{\partial \Delta_n} \left(f(z) - f(z_0) - (z-z_0)f'(z_0)\right)\,\mbox{d} z\right| \\ &\leqslant \operatorname{length}(\partial \Delta_n)\varepsilon\sup_{z \in \partial \Delta_n}|z-z_0|.\end{align} ``` The solution is not that easy, it is not trivial to chose which & remove for having the correct the alignment