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Getting the argand with De Moires theorem

  • Thread starter thomas49th
  • Start date
  • #1
655
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Hi just did an interesting question which made me wonder about solving it another way around

z = 2 + 5i
find the value of [/tex]z^{\frac{1}{4}}[/tex] which lies in the second quadrant of the Argand diagram and enter it's argument


So [tex]arctan(5/2) \approx 1.1902 [/tex]

We use De Moivre's Thereom and the the oscillation of the trigonometric functions
[tex]\sqrt{29}(cos(\frac{1.1902 + 2k\pi}{4}) + isin(\frac{1.1902 + 2k\pi}{4}))

So to lie in the second quadrant the angle must be [tex]\theta > \frac{\pi}{2}[/tex]
[tex]\theta > \frac{ \pi}{2}[/tex]
[tex]\frac{1.1902 + 2k\pi}{4} > \frac{\pi}{2}[/tex]

so k > 0.8 so k > 0 (where k is an integer)

The next integer after 0 is 1! So sub k = 1 into [tex]\frac{1.1902 + 2\pi}{4}[/tex]

then convert to degrees 107° agreed???

BUT how come if we say:
To lie in the second quadrant [tex]\theta < \pi[/tex]

[tex]\frac{1.1902 + 2k\pi}{4} < \pi[/tex]
solves k < 2.8 so if where an integer k can take the ranges 1 to 2.

BUT if we put 2 in:
[tex]\frac{1.1902 + 4\pi}{4}[/tex] this is greater than [tex]\pi[/tex]

Thanks
Thomas
 

Answers and Replies

  • #2
gabbagabbahey
Homework Helper
Gold Member
5,002
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To lie in the second quadrant [tex]\theta < \pi[/tex]

[tex]\frac{1.1902 + 2k\pi}{4} < \pi[/tex]
solves k < 2.8
No it doesn't, try solving this inequality again, and show your steps if you get the same result.
 
  • #3
655
0
im an idiot :blushing:
 

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