Algebraic confusion

Question

How does

$$\frac{1}{b^2i}\oint\frac{1}{z}\frac{dz}{\lbrack k + \frac{1}{2i}\left( z - \frac{1}{z}\right) \rbrack^2}$$

become

$$\frac{1}{b^2i}\oint\frac{1}{z}\frac{dz}{\lbrack \frac{1}{2iz} \left( z^2 + 2kiz - 1\right) \rbrack^2}$$

Is the book wrong or am I missing something?

The simplification is obtained by finding a common denominator for the expression inside brackets. — Michael Joyce, Feb 12 '16 at 07:02

score 2 · Accepted Answer · answered Feb 12 '16 at 07:03

2

Notice that $$k+\frac{1}{2i}(z-\frac{1}{z})=\frac{1}{2i}(2ki+z-\frac{1}{z})=\frac{1}{2i}(\frac{2kiz}{z}+\frac{z^2}{z}-\frac{1}{z})=\frac{1}{2iz}(z^2+2kzi-1)$$

answered Feb 12 '16 at 07:03

Zelos Malum

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Algebraic confusion

1 Answers1