Given a complex valued function can be written as f(z) = w = u(x,y) +iv(x,y), where w is the real part of w and v is the imaginary part of v.

pancha3 2021-09-14 Answered

1. Given a complex valued function can be written as \(f(z) = w = u(x,y) + iv(x,y)\), where w is the real part of w and v is the imaginary part of v. Using algebraic manipulation figure out what wu and v is if
\(\displaystyle{f{{\left({z}\right)}}}=\overline{{{z}}}{\left({1}-{e}^{{{i}{\left({z}+\overline{{{z}}}\right)}}}\right)}\)
Here \(\displaystyle\overline{{{z}}}\) denotes the complex conjugate of z.
2. Using the concept of limits figure out what the second derivative of \(\displaystyle{f{{\left({z}\right)}}}={z}{\left({1}-{z}\right)}\) is.
3. Use the theorems of Limits that have been discussed before to show that
\(\displaystyle\lim_{{{z}\rightarrow{1}-{i}}}{\left[{x}+{i}{\left({2}{x}+{y}\right)}\right]}={1}+{i}.{\left[\text{Hint: Use }\ {z}={x}+{i}{y}\right]}\)

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Expert Answer

aprovard
Answered 2021-09-15 Author has 21381 answers
Step 1 The given function is \(\displaystyle{f{{\left({z}\right)}}}=\overline{{{z}}}{\left({1}-{e}^{{{i}{\left({z}+\overline{{{z}}}\right)}}}\right)}\)
Substitute \(\displaystyle{z}={x}+{i}{y}\ \text{ and }\ \overline{{{z}}}={x}-{i}{y}\) in the above function and simplify as follows.
Step 2
\(\displaystyle{f{{\left({z}\right)}}}={\left({x}-{i}{y}\right)}{\left({1}-{e}^{{{i}{\left({x}+{i}{y}+{x}-{i}{y}\right)}}}\right)}\)
\(\displaystyle={\left({x}-{i}{y}\right)}{\left({1}-{e}^{{{i}{2}{x}}}\right)}\)
\(\displaystyle={x}-{i}{y}-{\left({x}-{i}{y}\right)}{e}^{{{i}{2}{x}}}\)
\(\displaystyle={x}-{i}{y}-{\left({x}-{i}{y}\right)}{\left({\cos{{\left({2}{x}\right)}}}+{i}{\sin{{\left({2}{x}\right)}}}\right)}\)
\(\displaystyle={x}-{i}{y}-{\left[{x}{\cos{{\left({2}{x}\right)}}}+{i}{x}{\sin{{\left({2}{x}\right)}}}-{i}{y}{\cos{{\left({2}{x}\right)}}}+{y}{\sin{{\left({2}{x}\right)}}}\right]}\)
\(\displaystyle={x}-{i}{y}-{x}{\cos{{\left({2}{x}\right)}}}-{i}{x}{\sin{{\left({2}{x}\right)}}}+{i}{y}{\cos{{\left({2}{x}\right)}}}-{y}{\sin{{\left({2}{x}\right)}}}\)
\(\displaystyle={\left[{x}-{x}{\cos{{\left({2}{x}\right)}}}-{y}{\sin{{\left({2}{x}\right)}}}\right]}+{i}{\left[-{y}-{x}{\sin{{\left({2}{x}\right)}}}+{y}{\cos{{\left({2}{x}\right)}}}\right]}\)
\(\displaystyle={u}{\left({x},{y}\right)}+{i}{v}{\left({x},{y}\right)}\)
Thus, \(\displaystyle{u}{\left({x},{y}\right)}={x}-{x}{\cos{{\left({2}{x}\right)}}}-{y}{\sin{{\left({2}{x}\right)}}}\ \text{ and }\ {v}{\left({x},{y}\right)}={y}{\cos{{\left({2}{x}\right)}}}-{y}-{x}{\sin{{\left({2}{x}\right)}}}\)
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