• Mar 20th 2013, 02:36 AM
szak1592
i1/2

and

z1/n

and

-9

where z=x+i y
• Mar 20th 2013, 02:43 AM
Prove It
Quote:

Originally Posted by szak1592
i1/2

and

z1/n

and

-9

where z=x+i y

\displaystyle \displaystyle \begin{align*} i^{\frac{1}{2}} &= \left( e^{i \frac{\pi}{2}} \right)^{\frac{1}{2}} \\ &= e^{i \frac{\pi}{4}} \end{align*}

and

\displaystyle \displaystyle \begin{align*} z^{\frac{1}{n}} &= \left( r\, e^{i\theta} \right)^{\frac{1}{n}} \\ &= r^{\frac{1}{n}}\,e^{i\frac{\theta}{n}} \end{align*}

and

$\displaystyle \displaystyle -9 = 9\,e^{ i \pi }$
• Apr 9th 2013, 06:13 AM
HallsofIvy
For general $\displaystyle x+ iy= re^{i\theta}$ with $\displaystyle r= \sqrt{x^2+ y^2}$, $\displaystyle \theta= arctan(y/x)$
For general $\displaystyle x+ iy= re^{i\theta}$ with $\displaystyle r= \sqrt{x^2+ y^2}$, $\displaystyle \theta= arctan(y/x)$
$\displaystyle \displaystyle \theta$ is only $\displaystyle \displaystyle \arctan{ \left( \frac{y}{x} \right) }$ if the complex number is in the first quadrant. Otherwise you need to evaluate the angle in the first quadrant and then transform.