A problem in complex analysis

Show that if $f$ is holomorphic, then $$\Delta|f(z)+z|^2=4|f'(z)+1|^2.$$

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There are two ways to deal with the modulus: either use the formula $|z|^2=z\bar{z}$ or separate $z$ into real and imaginary parts. For ease of calculations, we separate $f(z)+z$ into real and imaginary parts, having $$\Delta|f(z)+z|^2=\Delta\bigg((u+x)^2+(v+y)^2\bigg).$$ [Here $u,v$ are actually $u(x,y)$ and $v(x,y)$ respectively.] $$\partial_x \bigg((u+x)^2+(v+y)^2\bigg)=2(u+x)(u_x+1)+2(v+y)v_x\\ \partial_x^2 \bigg((u+x)^2+(v+y)^2\bigg)=2(u+x)u_{xx}+2(v+y)v_{xx}+2(u_x+1)^2+2v_x^2$$ Therefore, $$\Delta\bigg((u+x)^2+(v+y)^2\bigg)=2(u+x)(u_{xx}+u_{yy})+2(v+y)(v_{xx}+v_{yy})\\+2\bigg( (u_x+1)^2+v_x^2+u_y^2+(v_y+1)^2\bigg).$$ Since $f$ is holomorphic, we have $$u_{xx}+u_{yy}=0\\ v_{xx}+v_{yy}=0\\ u_x=v_y,\quad u_y=-v_x.$$ Finally, $$\Delta|f(z)+z|^2=4\bigg( (u_x+1)^2+u_y^2\bigg)=4|f'(z)+1|^2$$ since $f'(z)+1=u_x+1+iu_y$ and $|f'(z)+1|=(u_x+1)^2+u_y^2$.