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Thread: Principal Logarithm

  1. #1
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    Principal Logarithm

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

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

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

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

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