Getting the argand with De Moires theorem

[thomas49th]

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 $$arctan(5/2) \approx 1.1902$$

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}))<br /> <br /> So to lie in the second quadrant the angle must be $$\theta > \frac{\pi}{2}$$ <br /> $$\theta > \frac{ \pi}{2}$$<br /> $$\frac{1.1902 + 2k\pi}{4} > \frac{\pi}{2}$$<br /> <br /> so k > 0.8 so k > 0 (where k is an integer)<br /> <br /> The next integer after 0 is 1! So sub k = 1 into $$\frac{1.1902 + 2\pi}{4}$$<br /> <br /> then convert to degrees 107° agreed?<br /> <br /> <b>BUT how come if we say:</b><br /> To lie in the second quadrant $$\theta < \pi$$<br /> <br /> $$\frac{1.1902 + 2k\pi}{4} < \pi$$<br /> solves k < 2.8 so if where an integer k can take the ranges 1 to 2.<br /> <br /> BUT if we put 2 in:<br /> $$\frac{1.1902 + 4\pi}{4}$$ this is greater than $$\pi$$<br /> <br /> Thanks<br /> Thomas[/tex]

 

[gabbagabbahey]

To lie in the second quadrant $$\theta < \pi$$

$$\frac{1.1902 + 2k\pi}{4} < \pi$$
solves k < 2.8

No it doesn't, try solving this inequality again, and show your steps if you get the same result.

 

[thomas49th]

im an idiot