Hi just did an interesting question which made me wonder about solving it another way around(adsbygoogle = window.adsbygoogle || []).push({});

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

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# Homework Help: Getting the argand with De Moires theorem

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