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Question Number 1794 by Rasheed Soomro last updated on 29/Sep/15
Simplify log_((3+4i)) (−7+24i)
Simplifylog(3+4i)(−7+24i)
Commented by 112358 last updated on 29/Sep/15
Define p=log_((3+4i)) (−7+24i) By change of base to base e we can rewrite p as p=((ln(−7+24i))/(ln(3+4i))) . Let q=3+4i. In exponential form we get q=5e^(itan^(−1) (4/3)) so then lnq=ln5+itan^(−1) (4/3). Let w=−7+24i. The modulus of w is found to be ∣w∣=25. The principal argument θ of w is given by θ=π−tan^(−1) (((24)/7)). We then get in exponential form w=25e^(i(π−tan^(−1) (24/7))) . Therefore lnw=ln25+i(π−tan^(−1) (24/7)) Thus p=((ln25+i(π−tan^(−1) (24/7)))/(ln5+itan^(−1) (4/3))) p=(((2ln5+i(π−tan^(−1) [((24)/7)]))(ln5−itan^(−1) [(4/3)]))/((ln5)^2 +(tan^(−1) [(4/3)])^2 )) p=((2(ln5)^2 +(π−tan^(−1) [((24)/7)])tan^(−1) [(4/3)]+i{π−tan^(−1) [((24)/7)]−2(tan^(−1) [(4/3)])}ln5)/((ln5)^2 +(tan^(−1) [(4/3)])^2 )) From this one can extract the real and imaginary parts of p. So log_((3+4i)) (−7+24i)=x+iy where x=Re(p) and y=Im(p).
Definep=log(3+4i)(−7+24i)Bychangeofbasetobaseewecanrewritepasp=ln(−7+24i)ln(3+4i).Letq=3+4i.Inexponentialformwegetq=5eitan−1(4/3)sothenlnq=ln5+itan−1(4/3).Letw=−7+24i.Themodulusofwisfoundtobe∣w∣=25.Theprincipalargumentθofwisgivenbyθ=π−tan−1(247).Wethengetinexponentialformw=25ei(π−tan−1(24/7)).Thereforelnw=ln25+i(π−tan−1(24/7))Thusp=ln25+i(π−tan−1(24/7))ln5+itan−1(4/3)p=(2ln5+i(π−tan−1[247]))(ln5−itan−1[43])(ln5)2+(tan−1[43])2p=2(ln5)2+(π−tan−1[247])tan−1[43]+i{π−tan−1[247]−2(tan−1[43])}ln5(ln5)2+(tan−1[43])2Fromthisonecanextracttherealandimaginarypartsofp.Solog(3+4i)(−7+24i)=x+iywherex=Re(p)andy=Im(p).

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