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Question Number 148636 by tabata last updated on 29/Jul/21

find tylor series of f(z)=logz   about z_o =−1+i

$${find}\:{tylor}\:{series}\:{of}\:{f}\left({z}\right)={logz}\: \\ $$$${about}\:{z}_{{o}} =−\mathrm{1}+{i} \\ $$

Commented by tabata last updated on 29/Jul/21

????

$$???? \\ $$

Answered by mathmax by abdo last updated on 30/Jul/21

f(z)=Σ_(n=0) ^∞  ((f^((n)) (z_0 ))/(n!))(z−z_0 )^n   f(z)=lnz ⇒f^′ (z)=(1/z) ⇒f^((n)) (z)=(((−1)^(n−1) (n−1)!)/z^n ) ⇒  f(z)=f(z_0 )+Σ_(n=1) ^∞  (((−1)^(n−1) (n−1)!)/(n!z_0 ^n ))(z−z_0 )^n   =log(−1+i)+Σ_(n=1) ^∞  (((−1)^(n−1) )/(n(−1+i)^n ))(z+1−i)^n   =iπ+log(e^(−((iπ)/4)) )−Σ_(n=1) ^∞  (1/n)e^((inπ)/4) (z+1−i)^n   logz=((3iπ)/4)−Σ_(n=1) ^∞  (1/n)e^((inπ)/4) (z+1−i)^n

$$\mathrm{f}\left(\mathrm{z}\right)=\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{f}^{\left(\mathrm{n}\right)} \left(\mathrm{z}_{\mathrm{0}} \right)}{\mathrm{n}!}\left(\mathrm{z}−\mathrm{z}_{\mathrm{0}} \right)^{\mathrm{n}} \\ $$$$\mathrm{f}\left(\mathrm{z}\right)=\mathrm{lnz}\:\Rightarrow\mathrm{f}^{'} \left(\mathrm{z}\right)=\frac{\mathrm{1}}{\mathrm{z}}\:\Rightarrow\mathrm{f}^{\left(\mathrm{n}\right)} \left(\mathrm{z}\right)=\frac{\left(−\mathrm{1}\right)^{\mathrm{n}−\mathrm{1}} \left(\mathrm{n}−\mathrm{1}\right)!}{\mathrm{z}^{\mathrm{n}} }\:\Rightarrow \\ $$$$\mathrm{f}\left(\mathrm{z}\right)=\mathrm{f}\left(\mathrm{z}_{\mathrm{0}} \right)+\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{n}−\mathrm{1}} \left(\mathrm{n}−\mathrm{1}\right)!}{\mathrm{n}!\mathrm{z}_{\mathrm{0}} ^{\mathrm{n}} }\left(\mathrm{z}−\mathrm{z}_{\mathrm{0}} \right)^{\mathrm{n}} \\ $$$$=\mathrm{log}\left(−\mathrm{1}+\mathrm{i}\right)+\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{n}−\mathrm{1}} }{\mathrm{n}\left(−\mathrm{1}+\mathrm{i}\right)^{\mathrm{n}} }\left(\mathrm{z}+\mathrm{1}−\mathrm{i}\right)^{\mathrm{n}} \\ $$$$=\mathrm{i}\pi+\mathrm{log}\left(\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)−\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{\mathrm{n}}\mathrm{e}^{\frac{\mathrm{in}\pi}{\mathrm{4}}} \left(\mathrm{z}+\mathrm{1}−\mathrm{i}\right)^{\mathrm{n}} \\ $$$$\mathrm{logz}=\frac{\mathrm{3i}\pi}{\mathrm{4}}−\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{\mathrm{n}}\mathrm{e}^{\frac{\mathrm{in}\pi}{\mathrm{4}}} \left(\mathrm{z}+\mathrm{1}−\mathrm{i}\right)^{\mathrm{n}} \\ $$

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