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
\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 &amp;gt; \frac{\pi}{2} <br /> \theta &amp;gt; \frac{ \pi}{2}<br /> \frac{1.1902 + 2k\pi}{4} &amp;gt; \frac{\pi}{2}<br /> <br /> so k &gt; 0.8 so k &gt; 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 &amp;lt; \pi<br /> <br /> \frac{1.1902 + 2k\pi}{4} &amp;lt; \pi<br /> solves k &lt; 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

 

[gabbagabbahey]

To lie in the second quadrant \theta &lt; \pi

\frac{1.1902 + 2k\pi}{4} &lt; \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