CVXI1

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Give an upper bound for {\Bigg |}\int _{{|z|=3}}{\frac  {dz}{z^{2}-i}}{\Bigg |}\,.

Using the fact that {\Bigg |}\int _{\Gamma }f(z)dz{\Bigg |}\leq {\mathrm  {max}}_{{z\,{\mathrm  {on}}\,\Gamma }}|f(z)|\cdot {\mathrm  {length}}(\Gamma )\,, in this case

f(z)={\frac  {1}{z^{2}-i}}\,

{\Big |}{\frac  {1}{z^{2}-i}}{\Big |}={\frac  {1}{|z^{2}-i|}}\,

{\mathrm  {min}}_{{|z^{2}-i|}}=|9i-i|=|8i|=8\,

Therefore

{\mathrm  {max}}_{{z\in C}}|f(z)|={\frac  {1}{8}}\,

{\Bigg |}\int _{{|z|=3}}{\frac  {dz}{z^{2}-i}}{\Bigg |}\leq {\frac  {1}{8}}\int _{{|z|=3}}|dz|={\frac  {1}{8}}2\pi 3={\frac  {3\pi }{4}}\,


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