# Using the definition of complex derivative, evaluate f(z) expression using derivative operation based on limiting case as lim_{triangle z ->0}

3) Answer the following questions considering the complex functions given below.
a) Using the definition of complex derivative, evaluate f(z) expression using derivative operation based on limiting case as $$\displaystyle\lim_{{\triangle{z}\rightarrow{0}}}$$
a.1 $$\displaystyle{f{{\left({z}\right)}}}={\frac{{{1}}}{{{z}^{{2}}}}},{\left({z}\ne{}{0}\right)}$$

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Step 1
a.1 We have the definition complex derivative -
$$\displaystyle{f}{\left({z}\right)}=\lim_{{\triangle{z}\rightarrow{0}}}{\frac{{{f{{\left({z}+\triangle{z}\right)}}}-{f{{\left({z}\right)}}}}}{{\triangle{z}}}}$$
Given, $$\displaystyle{f{{\left({z}\right)}}}={\frac{{{1}}}{{{z}^{{2}}}}},{\left({z}\ne{}{0}\right)}$$
Then,
$$\displaystyle{f}{\left({z}\right)}=\lim_{{\triangle{z}\rightarrow{0}}}{\frac{{{\frac{{{1}}}{{{\left({z}+\triangle{z}\right)}^{{2}}}}}-{\frac{{{1}}}{{{z}^{{2}}}}}}}{{\triangle{z}}}}$$
$$\displaystyle=\lim_{{\triangle{z}\rightarrow{0}}}{\frac{{{z}^{{2}}-{\left({z}+\triangle{z}\right)}^{{2}}}}{{{\left({z}+\triangle{z}\right)}^{{2}}{z}^{{2}}\triangle{z}}}}$$
$$\displaystyle=\lim_{{\triangle{z}\rightarrow{0}}}{\frac{{{1}}}{{{z}^{{2}}}}}{\frac{{{0}-{2}{\left({z}+\triangle{z}\right)}{1}}}{{{\left({z}+\triangle{z}\right)}^{{21}}+\triangle{z}{z}{\left({z}+\triangle{z}\right)}{1}}}}$$
$$\displaystyle={\frac{{{1}}}{{{z}^{{2}}}}}\times{\frac{{-{2}{z}}}{{{z}^{{2}}+{0}}}}$$
$$\displaystyle={\frac{{{1}}}{{{z}^{{2}}}}}\times{\frac{{-{2}}}{{{z}}}}$$
$$\displaystyle=-{2}{z}^{{-{3}}}$$
$$\displaystyle\text{Hence, }\ {\left({\frac{{{1}}}{{{z}^{{2}}}}}\right)}^{{1}}=-{2}{z}^{{-{3}}},{z}\ne{}{0}$$
Step 2
Given, $$\displaystyle{f{{\left({z}\right)}}}={\frac{{{z}}}{{{z}+{1}}}},{\left({z}\ne{}-{1}\right)}$$
$$\displaystyle\text{Then, }\ {f}`{\left({z}\right)}=\lim_{{\triangle{z}\rightarrow{0}}}{\frac{{{\frac{{{z}+\triangle{z}}}{{{z}+\triangle{z}+{1}}}}-{\frac{{{z}}}{{{z}+{1}}}}}}{{\triangle{z}}}}$$
$$\displaystyle=\lim_{{\triangle{z}\rightarrow{0}}}{\frac{{{\left({z}+\triangle{z}\right)}{\left({z}+{1}\right)}-{z}{\left({z}+\triangle{z}+{1}\right)}}}{{{\left({z}+\triangle{z}+{1}\right)}{\left({z}+{1}\right)}\triangle{z}}}}\ \ {\left({\frac{{{0}}}{{{0}}}}\ \text{ form}\right)}$$
$$\displaystyle=\lim_{{\triangle{z}\rightarrow{0}}}{\frac{{{1}}}{{{z}+{1}}}}{\frac{{{\left({z}+{1}\right)}\cdot{1}-{z}\cdot{1}}}{{{\left({z}+\triangle{z}+{1}\right)}\cdot{1}+\triangle{z}\cdot{1}}}}$$
$$\displaystyle=\lim_{{\triangle{z}\rightarrow{0}}}{\frac{{{1}}}{{{z}+{1}}}}{\frac{{{z}+{1}-{z}}}{{{\left({z}+\triangle{z}+{1}\right)}+\triangle{z}}}}$$
$$\displaystyle={\frac{{{1}}}{{{\left({z}+{1}\right)}^{{2}}}}}$$
$$\displaystyle\text{Hence, }\ {\left({\frac{{{z}}}{{{z}+{1}}}}\right)}^{{1}}={\frac{{{1}}}{{{\left({z}+{1}\right)}^{{2}}}}},{z}\ne{}-{1}$$

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