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Math Help - Principal Logarithm

  1. #1
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    Principal Logarithm

    Show that for all points  z in the right half plane  x>0 the function  \text{Log} \ z can be written  \text{Log} \ z = \frac{1}{2} \text{Log} \ (x^2+y^2) + i \arctan \frac{y}{x} .

    So  \text{Log} \ z = \text{Log} \ r + i  \arg z . Now  r = |z| = \sqrt{x^2+y^2} . So  \text{Log} \ z = \frac{1}{2} \text{Log} \ (r^2) + i \arctan \frac{y}{x}  = \text{Log} \ r + i \arctan \frac{y}{x} where  \arg z = \arctan \frac{y}{x}.

    Is this correct?
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  2. #2
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    Quote Originally Posted by manjohn12 View Post
    Show that for all points  z in the right half plane  x>0 the function  \text{Log} \ z can be written  \text{Log} \ z = \frac{1}{2} \text{Log} \ (x^2+y^2) + i \arctan \frac{y}{x} .

    So  \text{Log} \ z = \text{Log} \ r + i \arg z . Now  r = |z| = \sqrt{x^2+y^2} . So  \text{Log} \ z = \frac{1}{2} \text{Log} \ (r^2) + i \arctan \frac{y}{x} = \text{Log} \ r + i \arctan \frac{y}{x} where  \arg z = \arctan \frac{y}{x}.

    Is this correct?
    Looks fine. Now, what would be the difference if the question asked for all points  z in the left half plane  x<0 ....
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  3. #3
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    If  x < 0 the magnitude still stays the same. So  \arg z changes, and we have  - \arctan \frac{y}{x} .
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