Question

Need help... Topic: Complex numbers.

Answer 1

*complex

Answer 2

(z + 2) = z - (-2)

Answer 3

when z = 5, does that imply you just plug in z = 5

Answer 4

n=5

Answer 5

sorry. I typed wrong

Answer 6

Here's the correct question

Answer 7

I might need a hint.

Answer 8

Me too.. I am not able to find the correct direction from which this sum can be started

Answer 9

If i plug it into wolfram i get 5 roots
http://www.wolframalpha.com/input/?i=solve+(z%2B2)%5E5+%2B+z%5E5+%3D+0

Answer 10

Which tells me z = -1 is one solution
which is located at (-1,0) on the complex plane

Answer 11

Also see roots have the same real part, that means they all lie on a straight line

Answer 12

Do you need an elegant solution? @owen3

Answer 13

Sure. I am not sure if the question is asking for a single answer.

Answer 14

yes jango, please.

Answer 15

z+2=(z+1)+1 and z=(z+1)-1. so when n=5, we have
{(z+1)+1}^5+{(z+1)-1}^5=0

Answer 16

Let z+1=p
then (p+1)^5+(p-1)^5=0
from here we need to find p, which is very easy,

Answer 17

$$p^5 + 5p^4 + 10p^3 + 10p^2 + 5p + 1 +
\\ + p^5 - 5p^4 + 10p^3 - 10p^2 + 5p - 1 =0

$$

Answer 18

no.. dont expand

Answer 19

oh solve as is

Answer 20

{(p+1)/(p-1)}^5=-1=(cos pi+i sin pi)

Answer 21

then (p+1)/(p-1)= cos ((2kpi+pi)/5)+i sin ((2kpi+pi)/5), k=0,1,2,3,4

Answer 22

Demoivre's theorem?

Answer 23

absolutely. :)

Answer 24

you still have a complicated expression on the left side of equation
$$ \frac{p+1}{p-1} = \cos \left( \frac{2\pi k}{5} + \frac{\pi}{5} \right) + i \sin \left( \frac{2\pi k}{5} + \frac{\pi}{5} \right) $$

Answer 25

use componendo and dividendo

Answer 26

https://brilliant.org/wiki/componendo-and-dividendo/

Answer 27

what i tried to say, if you use compoenedo and dividendo, the left hand side will become p

Answer 28

and in the right hand side, we will have to do something

Answer 29

hmm i dont see how compendendo does that , can you show me.

Answer 30

sure. :)

Answer 31

thanks

Answer 32

here comes the elegance

Answer 33

Question.
Should that say $$ w = \frac{\pi k}{5} + \frac{\pi}{10} $$

Answer 34

yeah

Answer 35

use needed to make p an imaginary number..

Answer 36

I'm confused on the compenendo and dividendo step.

Answer 37

it says that if a/b=c/d, then (a+b)/(a-b)=(c+d)/(c-d)

Answer 38

oh :)

Answer 39

Thats a remarkable simplification.

Answer 40

you used a trig identity
1 + cos 2w =

Answer 41

right. also 1-cos 2w and used -1= i^2

Answer 42

1 + cos(2w) = 2 (cos w)^2

Answer 43

correct

Answer 44

so we get
$$ \rm \large z = -1 - i \cot\left( \frac{\pi }{10} + \frac{\pi k }{5} \right) , ~~k = 0,1,2,3,4$$

Answer 45

correct... and the geometrical interpretation is that all the points lie on a straight line whose real part is -1.

Answer 46

I think i see the compenendo derivation now.

$$ \rm \large {
\text{given:} ~~ \frac{a}{b } = \frac c d
\\ \, \\ \text{compendendo:}
\\ \\ \, \\
\frac{a+b}{b} = \frac{c+d}{d} \implies \frac{a+b}{c+d}= \frac b d
\\ \, \\ \text{dividendo:}
\\ \\ \, \\
\frac{a-b}{b} = \frac{c-d}{d} \implies \frac{a-b}{c-d}= \frac b d
\\ \, \\

\\ \, \\
\therefore \frac{a+b}{c+d} = \frac{a-b}{c-d}

\\ \, \\
\\
\therefore
\\
\frac{a+b}{a-b} = \frac{c+d}{c-d}

} $$

Answer 47

Are you studying complex analysis?

Answer 48

It was one of my subjects last year