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Question-206637




Question Number 206637 by jshfnahdj last updated on 21/Apr/24
Answered by Frix last updated on 21/Apr/24
x^y  with x, y ∈C  We need x=re^(iθ)  and y=a+bi ⇒  x^y =(re^(iθ) )^(a+bi) =r^(a+bi) e^(iθ(a+bi)) =  =r^a r^(ib) e^(−bθ) e^(iaθ) =e^(aln r) e^(ibln r) e^(−bθ) e^(iaθ) =  =e^(aln r −bθ) e^(i(bln r +aθ)) =  =e^(aln r −bθ) (cos (bln r +aθ) +i sin (bln r +aθ))    1−3i=(√(10))e^(−tan^(−1)  3)   ⇒ (1−3i)^(−(3/2)i) =  =e^(−((3tan^(−1)  3)/2)) (cos ((3ln 10)/4) −i sin ((3ln 20)/4))≈  ≈−.0238822−.151706i
$${x}^{{y}} \:\mathrm{with}\:{x},\:{y}\:\in\mathbb{C} \\ $$$$\mathrm{We}\:\mathrm{need}\:{x}={r}\mathrm{e}^{\mathrm{i}\theta} \:\mathrm{and}\:{y}={a}+{b}\mathrm{i}\:\Rightarrow \\ $$$${x}^{{y}} =\left({r}\mathrm{e}^{\mathrm{i}\theta} \right)^{{a}+{b}\mathrm{i}} ={r}^{{a}+{b}\mathrm{i}} \mathrm{e}^{\mathrm{i}\theta\left({a}+{b}\mathrm{i}\right)} = \\ $$$$={r}^{{a}} {r}^{\mathrm{i}{b}} \mathrm{e}^{−{b}\theta} \mathrm{e}^{\mathrm{i}{a}\theta} =\mathrm{e}^{{a}\mathrm{ln}\:{r}} \mathrm{e}^{\mathrm{i}{b}\mathrm{ln}\:{r}} \mathrm{e}^{−{b}\theta} \mathrm{e}^{\mathrm{i}{a}\theta} = \\ $$$$=\mathrm{e}^{{a}\mathrm{ln}\:{r}\:−{b}\theta} \mathrm{e}^{\mathrm{i}\left({b}\mathrm{ln}\:{r}\:+{a}\theta\right)} = \\ $$$$=\mathrm{e}^{{a}\mathrm{ln}\:{r}\:−{b}\theta} \left(\mathrm{cos}\:\left({b}\mathrm{ln}\:{r}\:+{a}\theta\right)\:+\mathrm{i}\:\mathrm{sin}\:\left({b}\mathrm{ln}\:{r}\:+{a}\theta\right)\right) \\ $$$$ \\ $$$$\mathrm{1}−\mathrm{3i}=\sqrt{\mathrm{10}}\mathrm{e}^{−\mathrm{tan}^{−\mathrm{1}} \:\mathrm{3}} \\ $$$$\Rightarrow\:\left(\mathrm{1}−\mathrm{3i}\right)^{−\frac{\mathrm{3}}{\mathrm{2}}\mathrm{i}} = \\ $$$$=\mathrm{e}^{−\frac{\mathrm{3tan}^{−\mathrm{1}} \:\mathrm{3}}{\mathrm{2}}} \left(\mathrm{cos}\:\frac{\mathrm{3ln}\:\mathrm{10}}{\mathrm{4}}\:−\mathrm{i}\:\mathrm{sin}\:\frac{\mathrm{3ln}\:\mathrm{20}}{\mathrm{4}}\right)\approx \\ $$$$\approx−.\mathrm{0238822}−.\mathrm{151706i} \\ $$

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