Complex Number Help (1 Viewer)

[Green Yoda]

If
a=cos(a)+sin(a)i
b=cos(b)+sin(b)i
c=cos(c)+sin(c)i
and a+b+c=0
Prove that 1/a + 1/b +1/c =0

 

[calamebe]

Well all of the complex numbers a, b and c have modulus 1 right? So then 1/a, 1/b and 1/c will be their conjugates. See if you can go from there.

 

[Engomasters]

If
a=cos(a)+sin(a)i
b=cos(b)+sin(b)i
c=cos(c)+sin(c)i
and a+b+c=0
Prove that 1/a + 1/b +1/c =0

Heres the solution.

Bear in mind some steps may have been skipped.

 

[fluffchuck]

a + b + c = 0

cis(a) + cis(b) + cis(c) = 0

Now take the conjugate of both sides:

cis(-a) + cis(-b) + cis(-c) = 0

1/cis(a) + 1/cis(b) + 1/cis(c) = 0

1/a + 1/b + 1/c = 0

 

[jathu123]

an alternate method

oh lol nvm the above post is much shorter

 

[calamebe]

Heres the solution.

Bear in mind some steps may have been skipped.

Just letting you know you cannot use triangle formulas such as the cosine rule unless you are given that the complex numbers form the sides of a triangle, and even then you must use the modulus, not the complex number itself.

 

[calamebe]

Also guys, this results isn't even restricted to complex numbers with modulus 1.

Let a = rcis(α), b = rcis(β), c = rcis(θ)

If a + b + c = 0, then r(cis(α) + cis(β) + cis(θ)) = 0 or just cis(α) + cis(β) + cis(θ) = 0.

1/a + 1/b + 1/c = (cis(-α) + cis(-β) + cis(-θ))/r = 0 as again, cis(-α) + cis(-β) + cis(-θ) is the conjugate of cis(α) + cis(β) + cis(θ), and the conjugate of 0 is 0.