Author Topic: how to work out the conjugate (Read 1178 times) Tweet Share

1i1ii1i · « **on:** April 05, 2012, 01:30:52 pm »

how do you do the conjugate of -3cis(2pi/3) and 5cis(-pi/4)

Bhootnike · « **Reply #1 on:** April 05, 2012, 01:36:41 pm »

the conjugate of a complex number is $\bar{z} = x - yi = rcis(-\theta)$

thus it'd be ~~$-3cis(\frac{-2\pi}{3})$~~ and $5cis(\frac{\pi}{4})$

EDIT. look down

1i1ii1i · « **Reply #2 on:** April 05, 2012, 01:43:17 pm »

Yea so like that's what i did but that's wrong

1i1ii1i · « **Reply #3 on:** April 05, 2012, 01:43:43 pm »

the answer is for the first is 3cis(pi/3)

brightsky · « **Reply #4 on:** April 05, 2012, 02:16:46 pm »

-3cis(2pi/3) isn't really polar form, as the modulus is negative. you need to convert it to 3 cis(-pi/3)

Bhootnike · « **Reply #5 on:** April 05, 2012, 02:17:49 pm »

well thats awkward then..

mm , ok so.
$-3cis(\frac{2\pi}{3})$ = $-3(cos(\frac{2\pi}{3}) + i sin(\frac{2\pi}{3})$
$=-3(\frac{-1}{2} + i \frac{\sqrt{3}}{2}})$
$= \frac{-3}{2} + \frac{3\sqrt{3}}{2}}i$

thus conjugate = $= \frac{-3}{2} - \frac{3\sqrt{3}}{2}} i$

transforming back into polar form:
r = $\sqrt{\frac{-3}{2} ^2 + \frac{3\sqrt{3}}{2}}^2$
r= $\sqrt{\frac{9}{4} + \frac{27}{4}}$
r= $3$

$\theta = tan^-1 \frac{(\frac{3\sqrt{3}}{2}}{\frac{-3}{2}})$

$= tan^-1 ( -\sqrt{3}})$

thus, $\theta = \frac{-\pi}{3}$

so thus conjugate should be.. $3cis(\frac{-\pi}{3})$

1i1ii1i · « **Reply #6 on:** April 05, 2012, 03:56:06 pm »

thx alot bhoot nike makes much more sense now

Becky2012 · « **Reply #7 on:** April 05, 2012, 10:48:19 pm »

you can also use a diagram rather than converting back to cartesian form, remembering that a conjugate has simply been reflected across the x-axis. I usually find this much quicker and I'm less likely to make an arithmetic error

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Author Topic: how to work out the conjugate (Read 1178 times) Tweet Share

1i1ii1i

how to work out the conjugate

Bhootnike

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1i1ii1i

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1i1ii1i

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brightsky

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Bhootnike

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1i1ii1i

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Becky2012

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