Usando o ramo principal de z^{\lambda} calcule 2^{\sqrt{2}}, 12^{\sqrt{2}},(5 i)^{1+i}, 1^{i}, 1^{-i}.
(I) 2^{\sqrt{2}}=\exp \left(\sqrt{2} \log 2 e^{i 0}\right)=\exp (\sqrt{2} \log 2+\sqrt{2} i 0)=2^{\sqrt{2}}.
(II) 12^{\sqrt{2}}=12^{\sqrt{2}}.
(III) (5 i)^{1+i}=\exp ((1+i) \log (5 i))=\exp \left((1+i) \log |5 i|+(1+i) \arg _{0}(5 i)\right)=\exp \left((1+i) \log 5+(1+i) \frac{\pi}{2}\right)
=\exp \left(\frac{\pi}{2}+\log 5+i\left(\frac{\pi}{2}+\log 5\right)\right)=\exp \left(\frac{\pi}{2}+\log 5\right)\left(\cos \left(\frac{\pi}{2}+\log 5\right)+i \sin \left(\frac{\pi}{2}+\log 5\right)\right)
=5 e^{\pi / 2}\left(\cos \left(\frac{\pi}{2}+\log 5\right)+i \sin \left(\frac{\pi}{2}+\log 5\right)\right) .
(IV) 1^{ \pm i}=\exp [ \pm i \log 1]=\exp [ \pm i \cdot 0]=1.

Passo 1

O enunciado pede para calcular, no ramo principal de z^{\lambda}, os valores de
2^{\sqrt{2}},\quad 12^{\sqrt{2}},\quad (5i)^{1+i},\quad 1^{i},\quad 1^{-i}.
Para isso vamos usar a definição
z^{\lambda} \;=\; \exp\bigl(\lambda\,\Log z\bigr),
onde a função principal \Log z é dada por
\Log z \;=\;\ln|z| \;+\; i\,\Arg_{0}(z),
com \Arg_{0}(z)\in(-\pi,\pi].

Passo 2

Começando pelos casos de bases positivas reais, para z>0 temos \Arg_{0}(z)=0 e \Log z=\ln z. Logo
2^{\sqrt{2}} \;=\;\exp\bigl(\sqrt{2}\,\Log 2\bigr) =\exp\bigl(\sqrt{2}\,\ln2\bigr) =2^{\sqrt{2}},
que é simplesmente o número real 2^{\sqrt{2}}.

Passo 3

De modo análogo,
12^{\sqrt{2}} =\exp\bigl(\sqrt{2}\,\Log 12\bigr) =\exp\bigl(\sqrt{2}\,\ln12\bigr) =12^{\sqrt{2}}.

Passo 4

Agora (5i)^{1+i}. Primeiro note que
|5i|=5, \qquad \Arg_{0}(5i)=\tfrac{\pi}{2}.
Logo
\Log(5i)=\ln5 + i\tfrac{\pi}{2}.
Multiplicando por 1+i,
(1+i)\,\Log(5i) =(1+i)\bigl(\ln5 + i\tfrac{\pi}{2}\bigr) =\bigl(\ln5-\tfrac{\pi}{2}\bigr) +i\bigl(\ln5+\tfrac{\pi}{2}\bigr).
Portanto
(5i)^{1+i} =\exp\bigl((1+i)\Log(5i)\bigr) =\exp\bigl(\ln5-\tfrac{\pi}{2}\bigr)\, \exp\bigl(i(\ln5+\tfrac{\pi}{2})\bigr)
=5\,e^{-\tfrac{\pi}{2}}\Bigl[\cos\bigl(\ln5+\tfrac{\pi}{2}\bigr) +i\sin\bigl(\ln5+\tfrac{\pi}{2}\bigr)\Bigr].

Passo 5

Por fim, para 1^{\pm i} temos |1|=1 e \Arg_{0}(1)=0, logo \Log1=0. Assim
1^{i}=\exp\bigl(i\cdot0\bigr)=1, \qquad 1^{-i}=\exp\bigl(-i\cdot0\bigr)=1.

Resposta

2^{\sqrt{2}}=2^{\sqrt{2}} 12^{\sqrt{2}}=12^{\sqrt{2}}
(5i)^{1+i} =5\,e^{-\tfrac{\pi}{2}}\Bigl[\cos\bigl(\ln5+\tfrac{\pi}{2}\bigr) +i\sin\bigl(\ln5+\tfrac{\pi}{2}\bigr)\Bigr] 1^{i}=1 1^{-i}=1

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