Hi having problems interpreting some conditions as usual. It's usually the interpretation of the conditions that are letting me down. Anyhow, its to do with Cauchy's Integral Formula.

Heres the question followed by my solution:

(a)

Let: $\displaystyle

\begin{array}{*{20}c}

{\gamma _1 = \{ z \in C:\left| {z - 3i} \right| = 2\} ,} & {\gamma _2 = \{ z \in C:\left| z \right| = 5\} } \\

\end{array} $ be 2 positively oriented contours. Evaluate the following integrals clearly stating which results are being used:

$\displaystyle

\begin{array}{*{20}c}

{I_1 = \int\limits_{\gamma _1 } {\frac{z}

{{z^2 + 4}}dz,} } & {I_2 = \int\limits_{\gamma _2 } {\frac{z}

{{z^2 + 4}}} } \\

\end{array} dz$

(b)

Using Cauchy Riemann prove that the function h(z)=sin(Imz) is not differentiable at any point of the strip $\displaystyle \{ z:\frac{{ - \pi }}

{2} < {Im} z < \frac{\pi }

{2}\}$

solution to (a):

----------------------

for contour 1: $\displaystyle

{\gamma _1 = \{ z \in C:\left| {z - 3i} \right| = 2\} }$.I interpret this as the contour being centred at 3i and having radius 2?

So using the Cauchy Integral formula: $\displaystyle f(z_0 ) = \frac{1}

{{2\pi i}}\int\limits_\gamma {\frac{{f(z)}}

{{z - z_0 }}} dz$

using partial fractions we have,

$\displaystyle I_1 = \int\limits_{\gamma _1 (3i,2)} {\frac{z}

{{z^2 + 4}}dz} = \int\limits_{\gamma _1 (3i,2)} {\frac{A}

{{z + 2i}}dz} + \int\limits_{\gamma _1 (3i,2)} {\frac{B}

{{z - 2i}}dz} $

so: $\displaystyle z = A(z - 2i) + (z + 2i)$

$\displaystyle

A = \frac{1}

{2},B = \frac{1}

{2}$

so now we have:

$\displaystyle

I_1 = \int\limits_{\gamma _1 (3i,2)} {\frac{1}

{{2(z + 2i)}}dz} + \int\limits_{\gamma _1 (3i,2)} {\frac{1}

{{2(z - 2i)}}dz} {\text{ }} = {\text{ }}\frac{1}

{2}\int\limits_{\gamma _1 (3i,2)} {\frac{1}

{{(z + 2i)}}} + \frac{1}

{{(z - 2i)}}dz$

again, the cauchy integral formula says:

$\displaystyle f(z_0 ) = \frac{1}

{{2\pi i}}\int\limits_\gamma {\frac{{f(z)}}

{{z - z_0 }}} dz$

so we let f(z)=z,

so now we get

$\displaystyle

(2i + ( - 2i)) = \frac{1}

{{2\pi i}}\frac{1}

{2}\int\limits_{\gamma _1 (3i,2)} {\frac{1}

{{(z + 2i)}}} + \frac{1}

{{(z - 2i)}}dz$

THE LHS GOES TO 0 SO THE WHOLE INTEGRAL IS THEREFORE 0?

----------------------------------------------------------------------

For the 2nd integral: $\displaystyle {\gamma _2 = \{ z \in C:\left| z \right| = 5\} }$,I interpret the contour being centred at 0 and having radius 5?.

However, the actual partial fractions are still the same and the LHS is still (2i + (-2i) which still makes the overall integral go to 0. So how are the any different.

Does this basically back up the fact from the Cauchy-Goursat theorem that any closed contour has an integral of 0?

Or am i misinterpreting the contour conditions?

solution for (b):

--------------------------

any suggestions here? i understand the cauchy-riemann equations say $\displaystyle u_x=v_y$ and $\displaystyle v_x=-u_y$ so I have to probably show that. but dont understand how it applies to the actual question.

thanks a lot people