# Getting the argand with De Moires theorem

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
$$\sqrt{29}(cos(\frac{1.1902 + 2k\pi}{4}) + isin(\frac{1.1902 + 2k\pi}{4})) So to lie in the second quadrant the angle must be [tex]\theta > \frac{\pi}{2}$$
$$\theta > \frac{ \pi}{2}$$
$$\frac{1.1902 + 2k\pi}{4} > \frac{\pi}{2}$$

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

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

then convert to degrees 107° agreed???

BUT how come if we say:
To lie in the second quadrant $$\theta < \pi$$

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

BUT if we put 2 in:
$$\frac{1.1902 + 4\pi}{4}$$ this is greater than $$\pi$$

Thanks
Thomas

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gabbagabbahey
Homework Helper
Gold Member
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.

im an idiot 