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

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
#3
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I would like to see too.

Aug 15, 2020
#4
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The following works, but is a little tedious: 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.

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

Aug 15, 2020
edited by Melody  Aug 15, 2020
#6
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Yes, they're the same.  Your method is very neat!

Alan  Aug 15, 2020
#7
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Thanks very much Alan   Melody  Aug 15, 2020
#8
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Here's an alternative route.

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

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

Aug 15, 2020
#9
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That is really cool!    Thanks Tiggsy!!

Melody  Aug 15, 2020