+0  
 
+1
81
9
avatar+177 

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
avatar
-1

|(z + w)^6| = 124.

 Aug 15, 2020
 #2
avatar+177 
+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
avatar+110835 
0

I would like to see too.

 Aug 15, 2020
 #4
avatar+30947 
+3

The following works, but is a little tedious:

 

 Aug 15, 2020
 #5
avatar+110835 
+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??}\\

 Aug 15, 2020
edited by Melody  Aug 15, 2020
 #6
avatar+30947 
+1

Yes, they're the same.  Your method is very neat!

Alan  Aug 15, 2020
 #7
avatar+110835 
0

Thanks very much Alan   laugh laugh cool

Melody  Aug 15, 2020
 #8
avatar+140 
+3

Here's an alternative route.

 

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

 

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

 Aug 15, 2020
 #9
avatar+110835 
0

That is really cool!    Thanks Tiggsy!! 

Melody  Aug 15, 2020

37 Online Users