heat equation solving quadratic equation with complex numbers


by dp182
Tags: complex numbers, heat equation, quadratic equation, substitution
dp182
dp182 is offline
#1
Nov1-11, 12:54 AM
P: 22
1. The problem statement, all variables and given/known data
given that kλ2-ρcpuλ-ρcpωi=0
plug into the quadratic formula and get out an equation that looks like this

λ=α+iβγ√(1+iδ) where α,β,γ,and δ are in terms of ρ,cp,u,k, and ω

2. Relevant equations
(-b√b2-4ac)/2a
2-ρcpuλ-ρcpωi=0
λ=α+iβγ√(1+iδ)
3. The attempt at a solution
so I plugged it in and came out with
(ρcpu/2k)(ρcpu/2k)√1+(4kωi/ρcpu2)
so γ=(ρcpu/2k) and δ=(4kω/ρcpu2)
but I'm unable to make the first term α+iβ and help would be great
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Mark44
Mark44 is offline
#2
Nov1-11, 01:47 PM
Mentor
P: 21,067
Quote Quote by dp182 View Post
1. The problem statement, all variables and given/known data
given that kλ2-ρcpuλ-ρcpωi=0
plug into the quadratic formula and get out an equation that looks like this

λ=α+iβγ√(1+iδ) where α,β,γ,and δ are in terms of ρ,cp,u,k, and ω

2. Relevant equations
(-b√b2-4ac)/2a
2-ρcpuλ-ρcpωi=0
λ=α+iβγ√(1+iδ)
3. The attempt at a solution
so I plugged it in and came out with
(ρcpu/2k)(ρcpu/2k)√1+(4kωi/ρcpu2)
Should be λ= (ρcpu/2k)(ρcpu/2k)√1+(4kωi/ρcpu2) since you are solving the quadratic equation for λ. Note that I didn't check your work.

Quote Quote by dp182 View Post
so γ=(ρcpu/2k) and δ=(4kω/ρcpu2)
but I'm unable to make the first term α+iβ and help would be great
They're not asking you to make the first term α+iβ, just that you have an equation that has the form they give. In what you have, α = ρcpu/(2k) and β = 0.


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