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