# Convert into polar form of e raise to power i(theta), please help.

• Mar 20th 2013, 03:36 AM
szak1592
Convert into polar form of e raise to power i(theta), please help.
i1/2

and

z1/n

and

-9

where z=x+i y
• Mar 20th 2013, 03:43 AM
Prove It
Re: Convert into polar form of e raise to power i(theta), please help.
Quote:

Originally Posted by szak1592
i1/2

and

z1/n

and

-9

where z=x+i y

\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 \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 -9 = 9\,e^{ i \pi } " alt="\displaystyle -9 = 9\,e^{ i \pi } " />
• Apr 9th 2013, 07:13 AM
HallsofIvy
Re: Convert into polar form of e raise to power i(theta), please help.
For general $x+ iy= re^{i\theta}$ with $r= \sqrt{x^2+ y^2}$, $\theta= arctan(y/x)$
• Apr 10th 2013, 01:18 AM
Prove It
Re: Convert into polar form of e raise to power i(theta), please help.
Quote:

Originally Posted by HallsofIvy
For general $x+ iy= re^{i\theta}$ with $r= \sqrt{x^2+ y^2}$, $\theta= arctan(y/x)$

$\displaystyle \theta$ is only $\displaystyle \arctan{ \left( \frac{y}{x} \right) }$ if the complex number is in the first quadrant. Otherwise you need to evaluate the angle in the first quadrant and then transform.