Please see link, I'm wondering how you end with RHS of equation on second line of image. Please can explain the step and algebra manipulation used
Thanks
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How did they get to final answer, −iexp(−ma)/4a^3(1 + ma) – Arsenal123 Apr 23 '16 at 19:12
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Notice that $a^2 + z^2 = (z - ia)(z + ia)$ which in in term cancels one power of $(z-ia)^2$ in the numerator. First simplify $(z-ia)^2/(a^2+z^2)^2$. The rest is computing the differential and then the limit. – Maximilian Gerhardt Apr 23 '16 at 19:16
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Yep I had gone past that stage, however I'm still ending up with the wrong answer – Arsenal123 Apr 23 '16 at 19:18
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Did you apply the quotient rule correctly? – Triatticus Apr 23 '16 at 19:21
2 Answers
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You have the expression
$$\frac{(z-ia)^2e^{imz}}{(a^2+z^2)^2}$$
First we note that the roots of $a^2+z^2$ are $z = \pm ia$ so we have
$$\frac{(z-ia)^2}{(a^2+z^2)^2} = \frac{(z-ia)^2}{\left((z-ia)(z+ia)\right)^2}$$
$$= \frac{(z-ia)^2}{(z-ia)^2(z+ia)^2} = \frac{1}{(z+ia)^2}$$
Now what's left to do is to first compute
$$\frac{d}{dz}\left(\frac{e^{imz}}{(z+ia)^2} \right)$$
using the quotient rule $(u/v)'= (u'v - uv')/v^2 $ , we have
$$\frac{d}{dz}\left(\frac{e^{imz}}{(z+ia)^2} \right) = \frac{ime^{imz}\cdot(z+ia)^2 - e^{imz}\cdot 2(z+ia)}{(z+ia)^4} = e^{imz} \frac{im(z+ia)^2 - 2(z+ia)}{(z+ia)^4}$$
That's enough to set $z = ia$ now, so we have
$$e^{im(ia)} \frac{im(2ia)^2 -2(2ia)}{(2ia)^4} = e^{-ma} \frac{-im\cdot 4a^2- 4ia}{16a^4} = ie^{-ma} \frac{-ma-1}{4a^3} = -ie^{-ma}\frac{(1+ma)}{4a^3}$$
I guess you must have made a mistake during the simplifcation.
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Perhaps they did directly
$$\lim_{z\to ai}\left[\frac{(z-ai)^2e^{imz}}{(z-ai)^2(z+ai)^2}\right]'=\lim_{z\to ai}\frac{ime^{imz}(z+ai)-2e^{imz}}{(z+ai)^3}=\frac{-me^{-ma}2a-2e^{-ma}}{-8a^3i}$$
$$=\frac{e^{-ma}(ma+1)}{4a^3i}=-\frac{ie^{-ma}}{4a^3}(ma+1)$$
DonAntonio
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