Convert complex number to polar form?

**stupidoldman** · November 24th 2011, 12:58 PM

Well, I need help converting 1/(1+j*w*c*r) to polar form, where j is the complex number sqrt(-1) and w,c,r are just variables. Can't split up the denominator to make it a sum of fractions, nor does inverting helps.

This equation is from a series connection with a voltage source to a resistor and capacitor, finding the voltage source of the capacitor (the Vin is not included in the equation I gave, or else it would be Vin/(1+j*w*c*r)).This question is here because I don't need help with the phasor analysis part, but rather the complex conversion part.

**Plato** · November 24th 2011, 01:06 PM

Originally Posted by stupidoldman

Well, I need help converting 1/(1+j*w*c*r) to polar form, where j is the complex number sqrt(-1) and w,c,r are just variables. Can't split up the denominator to make it a sum of fractions, nor does inverting helps.

If $z$ is a complex number then $\frac{1}{z}=\frac{\overline{z}}{~|z|^2}$

**HallsofIvy** · November 25th 2011, 05:49 AM

I don't see why inverting would not help, if $z= re^{i\theta}$ then $\frac{1}{z}= \frac{1}{r}e^{-i\theta}$ . 1+ jwcr has $r= \sqrt{1+ w^2c^2r^2}$ , $\theta= arctan(wcr)$ .

**Soroban** · November 25th 2011, 06:53 AM

Hello, stupidoldman!

$\text{I need help converting }\frac{1}{1+wcr{\bf j}}\text{ to polar form,}$

. . $\text{where }{\bf j} = \sqrt{\text{-}1}\text{ and }w,c,r\text{ are constants.}$

Multiply numerator and denominator by the conjugate:

. $\frac{1}{1 + wcr{\bf j}} \cdot\frac{1- wcr{\bf j}}{1-wcr{\bf j}} \;=\;\frac{1-wcr{\bf j}}{1 + w^2c^2r^2} \;=\;\frac{1}{1+w^2c^2r^2} - \frac{2wcr}{1+w^2c^2r^2}{\bf j}$

Got it?

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