This-might-be-helpful-How-to-easily-calculate-z-1-z-2-with-z-1-z-2-C-Transform-z-1-re-i-and-z-2-a-bi-z-1-z-2-re-i-a-bi-z-1-z-2-r-a-bi-e-b-ia-z-1-z-2-r-a-e-

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Question Number 192580 by Frix last updated on 21/May/23

This might be helpful: How to easily calculate z_1 ^z_2 with z_1 , z_2 ∈C: Transform z_1 =re^(iθ) and z_2 =a+bi ⇒ z_1 ^z_2 =(re^(iθ) )^(a+bi) z_1 ^z_2 =r^(a+bi) e^(−bθ+iaθ) z_1 ^z_2 =r^a e^(−bθ) r^(bi) e^(iaθ) z_1 ^z_2 =(r^a /e^(bθ) )e^(i(aθ+bln r))

$$\mathrm{This}\:\mathrm{might}\:\mathrm{be}\:\mathrm{helpful}: \\ $$$$\mathrm{How}\:\mathrm{to}\:\mathrm{easily}\:\mathrm{calculate}\:{z}_{\mathrm{1}} ^{{z}_{\mathrm{2}} } \:\mathrm{with}\:{z}_{\mathrm{1}} ,\:{z}_{\mathrm{2}} \:\in\mathbb{C}: \\ $$$$\mathrm{Transform}\:{z}_{\mathrm{1}} ={r}\mathrm{e}^{\mathrm{i}\theta} \:\mathrm{and}\:{z}_{\mathrm{2}} ={a}+{b}\mathrm{i}\:\Rightarrow \\ $$$${z}_{\mathrm{1}} ^{{z}_{\mathrm{2}} } =\left({r}\mathrm{e}^{\mathrm{i}\theta} \right)^{{a}+{b}\mathrm{i}} \\ $$$${z}_{\mathrm{1}} ^{{z}_{\mathrm{2}} } ={r}^{{a}+{b}\mathrm{i}} \mathrm{e}^{−{b}\theta+\mathrm{i}{a}\theta} \\ $$$${z}_{\mathrm{1}} ^{{z}_{\mathrm{2}} } ={r}^{{a}} \mathrm{e}^{−{b}\theta} {r}^{{b}\mathrm{i}} \mathrm{e}^{\mathrm{i}{a}\theta} \\ $$$${z}_{\mathrm{1}} ^{{z}_{\mathrm{2}} } =\frac{{r}^{{a}} }{\mathrm{e}^{{b}\theta} }\mathrm{e}^{\mathrm{i}\left({a}\theta+{b}\mathrm{ln}\:{r}\right)} \\ $$

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This-might-be-helpful-How-to-easily-calculate-z-1-z-2-with-z-1-z-2-C-Transform-z-1-re-i-and-z-2-a-bi-z-1-z-2-re-i-a-bi-z-1-z-2-r-a-bi-e-b-ia-z-1-z-2-r-a-e-

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