Principal Logarithm

**manjohn12** · January 4th 2009, 03:30 AM

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?

**mr fantastic** · January 4th 2009, 02:28 PM

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 $x<0$ ....