Question Number 1794 by Rasheed Soomro last updated on 29/Sep/15
$${Simplify} \\ $$$${log}_{\left(\mathrm{3}+\mathrm{4}{i}\right)} \left(−\mathrm{7}+\mathrm{24}{i}\right) \\ $$
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).](Q1795.png)
$${Define}\:{p}={log}_{\left(\mathrm{3}+\mathrm{4}{i}\right)} \left(−\mathrm{7}+\mathrm{24}{i}\right) \\ $$$${By}\:{change}\:{of}\:{base}\:{to}\:{base}\:{e} \\ $$$${we}\:{can}\:{rewrite}\:{p}\:{as}\: \\ $$$$\:\:\:\:{p}=\frac{{ln}\left(−\mathrm{7}+\mathrm{24}{i}\right)}{{ln}\left(\mathrm{3}+\mathrm{4}{i}\right)}\:. \\ $$$${Let}\:{q}=\mathrm{3}+\mathrm{4}{i}.\:{In}\:{exponential}\:{form} \\ $$$${we}\:{get}\:{q}=\mathrm{5}{e}^{{itan}^{−\mathrm{1}} \left(\mathrm{4}/\mathrm{3}\right)} \:{so}\:{then} \\ $$$${lnq}={ln}\mathrm{5}+{itan}^{−\mathrm{1}} \left(\mathrm{4}/\mathrm{3}\right). \\ $$$${Let}\:{w}=−\mathrm{7}+\mathrm{24}{i}.\:{The}\:{modulus}\:{of} \\ $$$${w}\:{is}\:{found}\:{to}\:{be}\:\mid{w}\mid=\mathrm{25}.\:{The}\: \\ $$$${principal}\:{argument}\:\theta\:{of}\:{w}\:{is}\:{given} \\ $$$${by}\:\theta=\pi−{tan}^{−\mathrm{1}} \left(\frac{\mathrm{24}}{\mathrm{7}}\right).\:{We}\:{then}\:{get} \\ $$$${in}\:{exponential}\:{form}\: \\ $$$${w}=\mathrm{25}{e}^{{i}\left(\pi−{tan}^{−\mathrm{1}} \left(\mathrm{24}/\mathrm{7}\right)\right)} .\:{Therefore} \\ $$$${lnw}={ln}\mathrm{25}+{i}\left(\pi−{tan}^{−\mathrm{1}} \left(\mathrm{24}/\mathrm{7}\right)\right) \\ $$$${Thus} \\ $$$${p}=\frac{{ln}\mathrm{25}+{i}\left(\pi−{tan}^{−\mathrm{1}} \left(\mathrm{24}/\mathrm{7}\right)\right)}{{ln}\mathrm{5}+{itan}^{−\mathrm{1}} \left(\mathrm{4}/\mathrm{3}\right)} \\ $$$${p}=\frac{\left(\mathrm{2}{ln}\mathrm{5}+{i}\left(\pi−{tan}^{−\mathrm{1}} \left[\frac{\mathrm{24}}{\mathrm{7}}\right]\right)\right)\left({ln}\mathrm{5}−{itan}^{−\mathrm{1}} \left[\frac{\mathrm{4}}{\mathrm{3}}\right]\right)}{\left({ln}\mathrm{5}\right)^{\mathrm{2}} +\left({tan}^{−\mathrm{1}} \left[\frac{\mathrm{4}}{\mathrm{3}}\right]\right)^{\mathrm{2}} } \\ $$$${p}=\frac{\mathrm{2}\left({ln}\mathrm{5}\right)^{\mathrm{2}} +\left(\pi−{tan}^{−\mathrm{1}} \left[\frac{\mathrm{24}}{\mathrm{7}}\right]\right){tan}^{−\mathrm{1}} \left[\frac{\mathrm{4}}{\mathrm{3}}\right]+{i}\left\{\pi−{tan}^{−\mathrm{1}} \left[\frac{\mathrm{24}}{\mathrm{7}}\right]−\mathrm{2}\left({tan}^{−\mathrm{1}} \left[\frac{\mathrm{4}}{\mathrm{3}}\right]\right)\right\}{ln}\mathrm{5}}{\left({ln}\mathrm{5}\right)^{\mathrm{2}} +\left({tan}^{−\mathrm{1}} \left[\frac{\mathrm{4}}{\mathrm{3}}\right]\right)^{\mathrm{2}} } \\ $$$${From}\:{this}\:{one}\:{can}\:{extract}\:{the} \\ $$$${real}\:{and}\:{imaginary}\:{parts}\:{of}\:{p}. \\ $$$${So}\:{log}_{\left(\mathrm{3}+\mathrm{4}{i}\right)} \left(−\mathrm{7}+\mathrm{24}{i}\right)={x}+{iy} \\ $$$${where}\:{x}={Re}\left({p}\right)\:{and}\:{y}={Im}\left({p}\right). \\ $$