Let \(z = 2 e^{10 \pi i/21} \) and \(w = e^{\pi i/7}\). Then what is \( |(z+w)^6|,\)the magnitude of\((z+w)^6\)?

littlemixfan Aug 15, 2020

#1

#2**+1 **

thank you so much, but could you show me how you got the answer, rather than just the solution?

littlemixfan
Aug 15, 2020

#5**+3 **

Thanks Alan,

I also got 343 but i think my method was a bit unconventional and I wan't sure about the power of 6 bit.

This is how I did it.

(the angles in the pic are not to scale)

\(\beta = \frac{10\pi}{21},\qquad \alpha = \frac{\pi}{7}\)

To find the length of the resultant vector I used the cosine rule.

\(length^2=1^2+2^2-2*1*2*cos(\pi-\beta +\alpha)\\ length^2=5-4cos(\pi-(\beta -\alpha))\\ length^2=5+4cos(\beta -\alpha)\\ length^2=5+4cos(\frac{10\pi}{21} -\frac{3\pi}{21})\\ length^2=5+4cos(\frac{\pi}{3})\\ length^2=5+4cos(\frac{\pi}{3})\\ length^2=5+2\\ length^2=7\\ length^6=343\\ \text{So I think this means that } |(z+w)|^6=343\\ ~\\ \text{But I am asked for }|(z+w)^6| \\ \text{Alan gave the same answer, are they always the same??}\\\)

**Could you comment please Alan?**

LaTex:

length^2=1^2+2^2-2*1*2*cos(\pi-\beta +\alpha)\\

length^2=5-4cos(\pi-(\beta -\alpha))\\

length^2=5+4cos(\beta -\alpha)\\

length^2=5+4cos(\frac{10\pi}{21} -\frac{3\pi}{21})\\

length^2=5+4cos(\frac{\pi}{3})\\

length^2=5+4cos(\frac{\pi}{3})\\

length^2=5+2\\

length^2=7\\

length^6=343\\

\text{So I think this means that } |(z+w)|^6=343\\

~\\

\text{But I am asked for }|(z+w)^6| \\

\text{Alan gave the same answer, are they always the same??}\\

Melody Aug 15, 2020

#8**+3 **

Here's an alternative route.

\(\displaystyle z+w=2e^{10\pi\imath/21}+e^{\pi\imath/7}=e^{\pi\imath/7}(2e^{\pi\imath/3}+1)\\ =e^{\pi\imath/7}(2\cos(\pi/3)+2\imath\sin(\pi/3)+1)\\ =e^{\pi\imath/7}(2+\imath\sqrt{3}). \)

\(\displaystyle \mid z+w\mid=\mid 2+\imath\sqrt{3} \mid=\sqrt{7},\\ \text{so}\\ \mid z+w \mid^{6}=(\sqrt{7})^{6}=7^{3}=343. \)

Tiggsy Aug 15, 2020