Help it seems so easy...De Moivre's Theorem can be used to find reciprocals of complex numbers. Recall from algebra that the reciprocal of x is 1/x, which can be expressed as x-1. Use this fact, along with de Moivre's Theorem, to find the reciprocal of the number below.

(2sqrt3 + 2i)

Question

Help it seems so easy...De Moivre's Theorem can be used to find reciprocals of complex numbers. Recall from algebra that the reciprocal of x is 1/x, which can be expressed as x-1. Use this fact, along with de Moivre's Theorem, to find the reciprocal of the number below.

(2sqrt3 + 2i)

anonymous · Answer

so I did what you might expect... (2sqrt3-2i)^-1...  I have done it over and over and I am still getting it wrong.

fibonaccichick666 · Answer

https://www.youtube.com/watch?v=Sf9gEzcVZkU

anonymous · Answer

frist calculate r=sq 16=4

fibonaccichick666 · Answer

First, you may like that, second.... I thought De Moivre's only worked with natural numbers?

anonymous · Answer

z=(2 sqr 3-2i)^-1   ==>  arg (z)=pi/6

anonymous · Answer

lzl=4

anonymous · Answer

z=4( cos pi/6 +i sin pi/6) ==>

anonymous · Answer

yeah sorry i was watching the video... so yes i got up to there...

anonymous · Answer

where i got lost is when i did it to the negative 1 power

fibonaccichick666 · Answer

that's why I am wondering about your definition of demoivre's

anonymous · Answer

==>  z^-1= 4^-1 (cos pi/6+i sin pi/6)^-1

anonymous · Answer

i could be wrong it was just strategy since it explains how u get the reciprocal

anonymous · Answer

yes i did that

fibonaccichick666 · Answer

see, the definition here also excludes -1. http://web.pdx.edu/~caughman/Cindy%20501%20Final.pdf So I think you may need to put it into polar form then use the fact that $$[e^{i \theta}]^{-1}=e^{-i \theta}$$

anonymous · Answer

z^n=r^n (cos n teta  + i sin n teta)

fibonaccichick666 · Answer

^That is only for n which are natural

anonymous · Answer

ok... i dont know enough to have an opinion, so can someone guide me here?

fibonaccichick666 · Answer

I mean, they essentially just want you to say oh hey by comparison it has to be 1/z<--(in polar form though, then be like oh wait, I can't have a complex number on the bottom of a fraction so I have to fix that, but I don't know if that is mathematically legal

fibonaccichick666 · Answer

ah it is legal, ok page 17 on this link shows that form, check it out http://web.pdx.edu/~caughman/Cindy%20501%20Final.pdf

ganeshie8 · Answer

Not knowing enough is a good thing, it lets you experiment with things. But here there is nothing to experiment. You hav De Moivre's thm and you can just follow your nose and use it blindly

fibonaccichick666 · Answer

it also shows the derivation

ganeshie8 · Answer

http://gyazo.com/1a7de40b0d4bc1c9e45c7d711c8de365 continue from here

anonymous · Answer

thanks.. ok... so I did do this

anonymous · Answer

wait so why would it be positive

anonymous · Answer

if its in q4

ganeshie8 · Answer

apply De Moivre's thm  - kick that exponent to the angle

ganeshie8 · Answer

$$\frac{1}{4} \left[\cos (\pi/6) + i \sin(\pi/6)\right]^{-1}  = \frac{1}{4} \left[\cos (-\pi/6) + i \sin(-\pi/6)\right]  $$

fibonaccichick666 · Answer

- exponent doesn't mean - answer ie. $x^{-3}=\frac{1}{x^3}$

anonymous · Answer

yes i realized what i did wrong... but im talking about before... x was positive but once we got the angle we knew it was in q4

ganeshie8 · Answer

thats a good observation! and yes its like taking the conjugate

ganeshie8 · Answer

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