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Thread: complex roots

  1. #1
    Senior Member Danneedshelp's Avatar
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    complex roots

    a) Show that $\displaystyle e^{(r_{1}+r_{2})t}=e^{r_{1}t}e^{r_{2}t}$ for any complex number $\displaystyle r_{1}$ and $\displaystyle r_{2}$.

    b) Show that $\displaystyle \frac{d}{dt}e^{rt}=re^{rt}$.

    For (a) I am stuck because $\displaystyle e^{(r_{1}+r_{2})t}=e^{(\lambda+i\mu+\lambda-i\mu)t}=e^{2\lambda\\t}$. I am not sure where to go from here. Should I just consider to arbitrary positive complex number rather than conjugate roots? I am considering roots because that is all that is talked about in the section.

    For (b) do I just differentiate $\displaystyle e^{\lambda\\t}(cos(\mu\\t)+isin(\mu\\t)$ since $\displaystyle e^{(\lambda+i\mu)t}=e^{\lambda\\t}(cos(\mu\\t)+isi n(\mu\\t)$ or can I just look at the power series expansion of $\displaystyle e^{(\lambda+i\mu)t}$ and differentiate that?
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  2. #2
    MHF Contributor
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    Quote Originally Posted by Danneedshelp View Post
    a) Show that $\displaystyle e^{(r_{1}+r_{2})t}=e^{r_{1}t}e^{r_{2}t}$ for any complex number $\displaystyle r_{1}$ and $\displaystyle r_{2}$.

    b) Show that $\displaystyle \frac{d}{dt}e^{rt}=re^{rt}$.

    For (a) I am stuck because $\displaystyle e^{(r_{1}+r_{2})t}=e^{(\lambda+i\mu+\lambda-i\mu)t}=e^{2\lambda\\t}$. I am not sure where to go from here. Should I just consider to arbitrary positive complex number rather than conjugate roots? I am considering roots because that is all that is talked about in the section.

    For (b) do I just differentiate $\displaystyle e^{\lambda\\t}(cos(\mu\\t)+isin(\mu\\t)$ since $\displaystyle e^{(\lambda+i\mu)t}=e^{\lambda\\t}(cos(\mu\\t)+isi n(\mu\\t)$ or can I just look at the power series expansion of $\displaystyle e^{(\lambda+i\mu)t}$ and differentiate that?
    Part (a)
    Where does it say $\displaystyle r_1$ and $\displaystyle r_2$ are conjugates?

    $\displaystyle r_1=a\pm b\mathbf{i}$
    $\displaystyle r_2=c\pm d \mathbf{i}$

    $\displaystyle e^{((a+b\mathbf{i})+(c+d\mathbf{i}))t}=e^{(a+b\mat hbf{i})t+(c+d\mathbf{i})t}=e^{(a+b\mathbf{i})t}e^{ (c+d\mathbf{i})t}=e^{r_1t}e^{r_2t}$
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