1. ## 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?

2. Originally Posted by manjohn12
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 $\displaystyle x<0$ ....

3. If $\displaystyle x < 0$ the magnitude still stays the same. So $\displaystyle \arg z$ changes, and we have $\displaystyle - \arctan \frac{y}{x}$.