heat equation solving quadratic equation with complex numbersby dp182 Tags: complex numbers, heat equation, quadratic equation, substitution 

#1
Nov111, 12:54 AM

P: 22

1. The problem statement, all variables and given/known data
given that kλ^{2}ρc_{p}uλρc_{p}ωi=0 plug into the quadratic formula and get out an equation that looks like this λ=α+iβ±γ√(1+iδ) where α,β,γ,and δ are in terms of ρ,c_{p},u,k, and ω 2. Relevant equations (b±√b^{2}4ac)/2a kλ^{2}ρc_{p}uλρc_{p}ωi=0 λ=α+iβ±γ√(1+iδ) 3. The attempt at a solution so I plugged it in and came out with (ρc_{p}u/2k)±(ρc_{p}u/2k)√1+(4kωi/ρc_{p}u^{2}) so γ=(ρc_{p}u/2k) and δ=(4kω/ρc_{p}u^{2}) but I'm unable to make the first term α+iβ and help would be great 



#2
Nov111, 01:47 PM

Mentor
P: 20,962




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