is this right when (a+bi)^(c+di) =∣a+bi∣^(c+di) e^(i(c+di)arg(a+bi)) ? I had let arg(a+bi)= { ((tan^(−1) ((b/a))),(a≥0 and b≥0)),((π−tan^(−1) (−(b/a))),(a<0 and b≥0)),((−(π−tan^(−1) ((b/a)))),(a<0 and b<0)),((−tan^(−1) ((b/a))),(a≥0 and b<0)) :} before I solved it (a+bi)^(c+di) =∣a+bi∣^(c+di) e^(i(c+di)arg(a+bi)) =∣a+bi∣^c ∣a+bi∣^di e^(ic∙arg(a+bi)) e^(−d∙arg(a+bi)) =∣a+bi∣^c (c^di )^(ln∣a+bi∣) e^(ic∙arg(a+bi)) e^(−d∙arg(a+bi))


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Question Number 217101 by MathematicalUser2357 last updated on 01/Mar/25
$is this right when (a+bi)^(c+di) =∣a+bi∣^(c+di) e^(i(c+di)arg(a+bi)) ? I had let arg(a+bi)= { ((tan^(−1) ((b/a))),(a≥0 and b≥0)),((π−tan^(−1) (−(b/a))),(a<0 and b≥0)),((−(π−tan^(−1) ((b/a)))),(a<0 and b<0)),((−tan^(−1) ((b/a))),(a≥0 and b<0)) :} before I solved it (a+bi)^(c+di) =∣a+bi∣^(c+di) e^(i(c+di)arg(a+bi)) =∣a+bi∣^c ∣a+bi∣^di e^(ic∙arg(a+bi)) e^(−d∙arg(a+bi)) =∣a+bi∣^c (c^di )^(ln∣a+bi∣) e^(ic∙arg(a+bi)) e^(−d∙arg(a+bi))$
$$\mathrm{is}\:\mathrm{this}\:\mathrm{right}\:\mathrm{when}\:\left({a}+{bi}\right)^{{c}+{di}} =\mid{a}+{bi}\mid^{{c}+{di}} {e}^{{i}\left({c}+{di}\right)\mathrm{arg}\left({a}+{bi}\right)} ? \\ $$$$\mathrm{I}\:\mathrm{had}\:\mathrm{let}\:\mathrm{arg}\left({a}+{bi}\right)=\begin{cases}{\mathrm{tan}^{−\mathrm{1}} \left(\frac{{b}}{{a}}\right)}&{{a}\geqslant\mathrm{0}\:\mathrm{and}\:{b}\geqslant\mathrm{0}}\\{\pi−\mathrm{tan}^{−\mathrm{1}} \left(−\frac{{b}}{{a}}\right)}&{{a}<\mathrm{0}\:\mathrm{and}\:{b}\geqslant\mathrm{0}}\\{−\left(\pi−\mathrm{tan}^{−\mathrm{1}} \left(\frac{{b}}{{a}}\right)\right)}&{{a}<\mathrm{0}\:\mathrm{and}\:{b}<\mathrm{0}}\\{−\mathrm{tan}^{−\mathrm{1}} \left(\frac{{b}}{{a}}\right)}&{{a}\geqslant\mathrm{0}\:\mathrm{and}\:{b}<\mathrm{0}}\end{cases}\:\mathrm{before}\:\mathrm{I}\:\mathrm{solved}\:\mathrm{it} \\ $$$$\left({a}+{bi}\right)^{{c}+{di}} =\mid{a}+{bi}\mid^{{c}+{di}} {e}^{{i}\left({c}+{di}\right)\mathrm{arg}\left({a}+{bi}\right)} \\ $$$$=\mid{a}+{bi}\mid^{{c}} \mid{a}+{bi}\mid^{{di}} {e}^{{ic}\centerdot\mathrm{arg}\left({a}+{bi}\right)} {e}^{−{d}\centerdot\mathrm{arg}\left({a}+{bi}\right)} \\ $$$$=\mid{a}+{bi}\mid^{{c}} \left({c}^{{di}} \right)^{\mathrm{ln}\mid{a}+{bi}\mid} {e}^{{ic}\centerdot\mathrm{arg}\left({a}+{bi}\right)} {e}^{−{d}\centerdot\mathrm{arg}\left({a}+{bi}\right)} \\ $$
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