Question Number 9661 by geovane10math last updated on 23/Dec/16
$$\left.{a}\right)\:\mathrm{2}^{{i}} \:=\: \\ $$$$ \\ $$$$\left.{b}\right)\:\left({a}_{\mathrm{1}} \:+\:{b}_{\mathrm{1}} {i}\right)^{{a}_{\mathrm{2}} \:+\:{b}_{\mathrm{2}} {i}} \:=\: \\ $$$${Powers}\:{of}\:{complex}\:{numbers}\:??? \\ $$
Commented by prakash jain last updated on 23/Dec/16
$${a}^{{x}} ={e}^{{x}\mathrm{ln}\:{a}} \\ $$$$\mathrm{2}^{{i}} ={e}^{{i}\mathrm{ln}\:\mathrm{2}} =\mathrm{cos}\left(\:\mathrm{ln}\:\mathrm{2}\right)+{i}\mathrm{sin}\:\left(\mathrm{ln}\:\mathrm{2}\right) \\ $$
Commented by FilupSmith last updated on 23/Dec/16
$$\left({a}+{bi}\right)^{{x}+{yi}} \\ $$$$\left({a}+{bi}\right)^{{x}+{yi}} =\left({a}+{bi}\right)^{{x}} \left({a}+{bi}\right)^{{yi}} \\ $$$$=\left({a}+{bi}\right)^{{x}} {e}^{{yi}\mathrm{ln}\left({a}+{bi}\right)} \\ $$$$=\left({a}+{bi}\right)^{{x}} \left\{\mathrm{cos}\left({y}\mathrm{ln}\left\{{a}+{bi}\right\}\right)+{i}\mathrm{sin}\left({y}\mathrm{ln}\left\{{a}+{bi}\right\}\right)\right\} \\ $$$${z}={a}+{bi}\:\:\:\:\Rightarrow\:\:\:{z}^{{x}+{yi}} ={z}^{{x}} \left(\mathrm{cos}\left({y}\mathrm{ln}\left({z}\right)\right)+{i}\mathrm{sin}\left({y}\mathrm{ln}\left({z}\right)\right)\right) \\ $$
Commented by geovane10math last updated on 23/Dec/16
$${Thanks} \\ $$