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For a local gauge transformation $\delta \phi = i \alpha(x)\phi$, and $\delta \phi^*=-i\alpha(x)\phi$.

These equations imply $\delta(\partial_{\mu}\phi)=i(\partial_{\mu} \alpha)\phi+i \alpha (\partial_{\mu}\phi)$ and $\delta(\partial_{\mu}\phi^*)=-i(\partial_{\mu} \alpha)\phi-i \alpha (\partial_{\mu}\phi)$$\delta(\partial_{\mu}\phi^*)=-i(\partial_{\mu} \alpha)\phi^*-i \alpha (\partial_{\mu}\phi^*)$

Therefore, $\delta \mathcal{L}=\delta(\partial_{\mu}\phi \partial^{\mu}\phi^*)-m^2 \delta(\phi* \phi)=\partial_{\mu}J^{\mu}$$\delta \mathcal{L}=\delta(\partial_{\mu}\phi \partial^{\mu}\phi^*)-m^2 \delta(\phi* \phi)=\partial_{\mu}\alpha J^{\mu}$, where $J^{\mu}=i(\phi \partial^{\mu}\phi^*-\phi^* \partial^{\mu}\phi)$

Setting $\delta S=\int \delta \mathcal{L}=0$, one gets $\partial_{\mu}J^{\mu}=0$

For a local gauge transformation $\delta \phi = i \alpha(x)\phi$, and $\delta \phi^*=-i\alpha(x)\phi$.

These equations imply $\delta(\partial_{\mu}\phi)=i(\partial_{\mu} \alpha)\phi+i \alpha (\partial_{\mu}\phi)$ and $\delta(\partial_{\mu}\phi^*)=-i(\partial_{\mu} \alpha)\phi-i \alpha (\partial_{\mu}\phi)$

Therefore, $\delta \mathcal{L}=\delta(\partial_{\mu}\phi \partial^{\mu}\phi^*)-m^2 \delta(\phi* \phi)=\partial_{\mu}J^{\mu}$, where $J^{\mu}=i(\phi \partial^{\mu}\phi^*-\phi^* \partial^{\mu}\phi)$

Setting $\delta S=\int \delta \mathcal{L}=0$, one gets $\partial_{\mu}J^{\mu}=0$

For a local gauge transformation $\delta \phi = i \alpha(x)\phi$, and $\delta \phi^*=-i\alpha(x)\phi$.

These equations imply $\delta(\partial_{\mu}\phi)=i(\partial_{\mu} \alpha)\phi+i \alpha (\partial_{\mu}\phi)$ and $\delta(\partial_{\mu}\phi^*)=-i(\partial_{\mu} \alpha)\phi^*-i \alpha (\partial_{\mu}\phi^*)$

Therefore, $\delta \mathcal{L}=\delta(\partial_{\mu}\phi \partial^{\mu}\phi^*)-m^2 \delta(\phi* \phi)=\partial_{\mu}\alpha J^{\mu}$, where $J^{\mu}=i(\phi \partial^{\mu}\phi^*-\phi^* \partial^{\mu}\phi)$

Setting $\delta S=\int \delta \mathcal{L}=0$, one gets $\partial_{\mu}J^{\mu}=0$

2 added 7 characters in body
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For a local gauge transformation $\delta \phi = i \alpha(x)\phi$, and $\phi^*=-i\alpha(x)\phi$$\delta \phi^*=-i\alpha(x)\phi$.

These equations imply $\delta(\partial_{\mu}\phi)=i(\partial_{\mu} \alpha)\phi+i \alpha (\partial_{\mu}\phi)$ and $\delta(\partial_{\mu}\phi^*)=-i(\partial_{\mu} \alpha)\phi-i \alpha (\partial_{\mu}\phi)$

Therefore, $\delta \mathcal{L}=\delta(\partial_{\mu}\phi \partial^{\mu}\phi^*-m^2 \delta(\phi* \phi))=\partial_{\mu}J^{\mu}$$\delta \mathcal{L}=\delta(\partial_{\mu}\phi \partial^{\mu}\phi^*)-m^2 \delta(\phi* \phi)=\partial_{\mu}J^{\mu}$, where $J^{\mu}=i(\phi \partial^{\mu}\phi^*)-\phi^* \partial^{\mu}\phi$$J^{\mu}=i(\phi \partial^{\mu}\phi^*-\phi^* \partial^{\mu}\phi)$

Setting $\delta S=\int \delta \mathcal{L}=0$, one gets $\partial_{\mu}J^{\mu}=0$

For a local gauge transformation $\delta \phi = i \alpha(x)\phi$, and $\phi^*=-i\alpha(x)\phi$.

These equations imply $\delta(\partial_{\mu}\phi)=i(\partial_{\mu} \alpha)\phi+i \alpha (\partial_{\mu}\phi)$ and $\delta(\partial_{\mu}\phi^*)=-i(\partial_{\mu} \alpha)\phi-i \alpha (\partial_{\mu}\phi)$

Therefore, $\delta \mathcal{L}=\delta(\partial_{\mu}\phi \partial^{\mu}\phi^*-m^2 \delta(\phi* \phi))=\partial_{\mu}J^{\mu}$, where $J^{\mu}=i(\phi \partial^{\mu}\phi^*)-\phi^* \partial^{\mu}\phi$

Setting $\delta S=\int \delta \mathcal{L}=0$, one gets $\partial_{\mu}J^{\mu}=0$

For a local gauge transformation $\delta \phi = i \alpha(x)\phi$, and $\delta \phi^*=-i\alpha(x)\phi$.

These equations imply $\delta(\partial_{\mu}\phi)=i(\partial_{\mu} \alpha)\phi+i \alpha (\partial_{\mu}\phi)$ and $\delta(\partial_{\mu}\phi^*)=-i(\partial_{\mu} \alpha)\phi-i \alpha (\partial_{\mu}\phi)$

Therefore, $\delta \mathcal{L}=\delta(\partial_{\mu}\phi \partial^{\mu}\phi^*)-m^2 \delta(\phi* \phi)=\partial_{\mu}J^{\mu}$, where $J^{\mu}=i(\phi \partial^{\mu}\phi^*-\phi^* \partial^{\mu}\phi)$

Setting $\delta S=\int \delta \mathcal{L}=0$, one gets $\partial_{\mu}J^{\mu}=0$

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For a local gauge transformation $\delta \phi = i \alpha(x)\phi$, and $\phi^*=-i\alpha(x)\phi$.

These equations imply $\delta(\partial_{\mu}\phi)=i(\partial_{\mu} \alpha)\phi+i \alpha (\partial_{\mu}\phi)$ and $\delta(\partial_{\mu}\phi^*)=-i(\partial_{\mu} \alpha)\phi-i \alpha (\partial_{\mu}\phi)$

Therefore, $\delta \mathcal{L}=\delta(\partial_{\mu}\phi \partial^{\mu}\phi^*-m^2 \delta(\phi* \phi))=\partial_{\mu}J^{\mu}$, where $J^{\mu}=i(\phi \partial^{\mu}\phi^*)-\phi^* \partial^{\mu}\phi$

Setting $\delta S=\int \delta \mathcal{L}=0$, one gets $\partial_{\mu}J^{\mu}=0$