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# Having troubles (0.72∠-5π/36)/(1.38∠-17π/90)

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Having troubles (0.72∠-5π/36)/(1.38∠-17π/90)

Guest Nov 2, 2015

#9
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Thanks for pointing out the mistake guys - I've now corrected it.

Alan  Nov 7, 2015
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#1
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Deleted

Melody  Nov 2, 2015
edited by Melody  Nov 2, 2015
#2
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(0.72×(-5)×pi/36)/(1.38×(-17)×pi/90)

=150/391

Guest Nov 2, 2015
#3
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They are complex numbers, or rather standard abbreviations for complex numbers.

0.72$$\angle (-5\pi/36)$$ is a complex number with modulus 0.72 and argument $$-5\pi/36$$.

The rule for division, (deduce it by writing the numbers in their cos + i.sin form), is divide the moduli and subtract the arguments.

So, for example $$12\angle(\pi/4)/2\angle(\pi/6)=6\angle(\pi/12).$$

Guest Nov 2, 2015
#4
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Silly me,  thank you :))

I  would like more explanation of the division please :)

Melody  Nov 2, 2015
#5
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Put the numbers in their polar forms, $$r(\cos\theta+\imath\sin\theta)$$, multiply top and bottom by the conjugate of the denominator and then, having multiplied out the brackets,  use some standard trig identities to simplify.

I have to go out now, I'll get back to it later if it's still a problem.

Guest Nov 2, 2015
#6
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Or write it in the form

$$r_1e^{i\theta_1}/r_2e^{i\theta_2}\rightarrow r_1e^{i\theta_1}\times \frac{1}{r_2}e^{-i\theta_2}\rightarrow \frac{r_1}{r_2}e^{i(\theta_1-\theta_2)}$$

Edited to correct the magnitudes.

Alan  Nov 2, 2015
edited by Alan  Nov 7, 2015
#7
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Shoudn't that be r1 divided by r2 ?

Guest Nov 2, 2015
#8
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yeah he raised the e^i theta to the power of negative 1 to get it on the numerator but forgot to do the same to r2.

Guest Nov 7, 2015
edited by Guest  Nov 7, 2015
#9
+26328
+5

Thanks for pointing out the mistake guys - I've now corrected it.

Alan  Nov 7, 2015
#10
+91024
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Thanks guys  :))

Melody  Nov 7, 2015

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