1. ## 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?

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

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