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

**i**^{1/2}

and

**z**^{1/n
and }

**-9**

*where z=x+i y*

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

Quote:

Originally Posted by

**szak1592** **i**^{1/2}

and

**z**^{1/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 } $

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

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

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

Quote:

Originally Posted by

**HallsofIvy** 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.