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i-i-e-pi-2-and-we-can-renote-complex-number-i-as-0-1-1-0-i-i-0-1-1-0-0-1-1-0-But-why-Matrix-Exponent-Calculate-Dosen-t-defined-I-mean-A

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Question Number 220190 by SdC355 last updated on 07/May/25
i^i =e^(−(π/2)) and we can renote complex number i as ((0,(−1)),(1,( 0)) ) i^i = ((0,(−1)),(1,( 0)) )^ ((0,(−1)),(1,( 0)) ) But why Matrix Exponent Calculate Dosen′t defined?? I mean A,B∈mat(m,m) why A^B dosen′t defined??
$${i}^{{i}} ={e}^{−\frac{\pi}{\mathrm{2}}} \: \\ $$$$\mathrm{and}\:\mathrm{we}\:\mathrm{can}\:\mathrm{renote}\:\mathrm{complex}\:\mathrm{number}\:\boldsymbol{{i}}\:\mathrm{as}\:\begin{pmatrix}{\mathrm{0}}&{−\mathrm{1}}\\{\mathrm{1}}&{\:\:\:\:\mathrm{0}}\end{pmatrix} \\ $$$$\boldsymbol{{i}}^{\boldsymbol{{i}}} =\begin{pmatrix}{\mathrm{0}}&{−\mathrm{1}}\\{\mathrm{1}}&{\:\:\:\:\mathrm{0}}\end{pmatrix}^{\begin{pmatrix}{\mathrm{0}}&{−\mathrm{1}}\\{\mathrm{1}}&{\:\:\:\:\mathrm{0}}\end{pmatrix}} \: \\ $$$$\:\mathrm{But}\:\mathrm{why}\:\mathrm{Matrix}\:\mathrm{Exponent}\:\mathrm{Calculate}\:\mathrm{Dosen}'\mathrm{t}\:\mathrm{defined}?? \\ $$$$\:\mathrm{I}\:\mathrm{mean}\:{A},{B}\in\mathrm{mat}\left({m},{m}\right) \\ $$$$\mathrm{why}\:\mathrm{A}^{\mathrm{B}} \:\mathrm{dosen}'\mathrm{t}\:\mathrm{defined}?? \\ $$
Answered by Ghisom last updated on 07/May/25
because: z_1 =a+bi, z_2 =c+di ⇒ z_1 ^z_2 =(a+bi)^(c+di) is not easy to calculate instead we need z_1 =re^(iθ) , z_2 =c+di, r>0∧−π<θ<π ⇒ z_1 ^z_2 =(re^(iθ) )^(c+di) =e^(cln r −dθ) e^((dln r +cθ)i) using matrices (((rcos θ),(−rsin θ)),((rsin θ),(rcos θ)) )^ ((c,(−d)),(d,c) ) = = (((e^(cln r −dθ) cos (dln r +cθ)),(−e^(cln r −dθ) sin (dln r +cθ))),((e^(cln r −dθ) sin (dln r +cθ)),(e^(cln r −dθ) cos (dln r +cθ))) ) doesn′t make any sense...
$$\mathrm{because}: \\ $$$${z}_{\mathrm{1}} ={a}+{b}\mathrm{i},\:{z}_{\mathrm{2}} ={c}+{d}\mathrm{i}\:\Rightarrow\:{z}_{\mathrm{1}} ^{{z}_{\mathrm{2}} } =\left({a}+{b}\mathrm{i}\right)^{{c}+{d}\mathrm{i}} \\ $$$$\mathrm{is}\:\mathrm{not}\:\mathrm{easy}\:\mathrm{to}\:\mathrm{calculate} \\ $$$$\mathrm{instead}\:\mathrm{we}\:\mathrm{need} \\ $$$${z}_{\mathrm{1}} ={r}\mathrm{e}^{\mathrm{i}\theta} ,\:{z}_{\mathrm{2}} ={c}+{d}\mathrm{i},\:{r}>\mathrm{0}\wedge−\pi<\theta<\pi \\ $$$$\Rightarrow \\ $$$${z}_{\mathrm{1}} ^{{z}_{\mathrm{2}} } =\left({r}\mathrm{e}^{\mathrm{i}\theta} \right)^{{c}+{d}\mathrm{i}} =\mathrm{e}^{{c}\mathrm{ln}\:{r}\:−{d}\theta} \mathrm{e}^{\left({d}\mathrm{ln}\:{r}\:+{c}\theta\right)\mathrm{i}} \\ $$$$\mathrm{using}\:\mathrm{matrices} \\ $$$$\begin{pmatrix}{{r}\mathrm{cos}\:\theta}&{−{r}\mathrm{sin}\:\theta}\\{{r}\mathrm{sin}\:\theta}&{{r}\mathrm{cos}\:\theta}\end{pmatrix}^{\begin{pmatrix}{{c}}&{−{d}}\\{{d}}&{{c}}\end{pmatrix}} = \\ $$$$=\begin{pmatrix}{\mathrm{e}^{{c}\mathrm{ln}\:{r}\:−{d}\theta} \mathrm{cos}\:\left({d}\mathrm{ln}\:{r}\:+{c}\theta\right)}&{−\mathrm{e}^{{c}\mathrm{ln}\:{r}\:−{d}\theta} \mathrm{sin}\:\left({d}\mathrm{ln}\:{r}\:+{c}\theta\right)}\\{\mathrm{e}^{{c}\mathrm{ln}\:{r}\:−{d}\theta} \mathrm{sin}\:\left({d}\mathrm{ln}\:{r}\:+{c}\theta\right)}&{\mathrm{e}^{{c}\mathrm{ln}\:{r}\:−{d}\theta} \mathrm{cos}\:\left({d}\mathrm{ln}\:{r}\:+{c}\theta\right)}\end{pmatrix} \\ $$$$\mathrm{doesn}'\mathrm{t}\:\mathrm{make}\:\mathrm{any}\:\mathrm{sense}… \\ $$
Commented by SdC355 last updated on 10/May/25
Thx
$$\mathrm{Thx} \\ $$

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