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a fost răspuns

Să se calculeze:
a) (1+i) la a8a, (1+i) la 36, )2+5i) la a2+(2-5i) la a2a

Răspuns :

[tex](1+i)^8=[(1+i)^2]^4=(1+2i+i^2)^4=(2i)^4=16i^4=16\\ (1+i)^{36}=[(1+i)^2]^{18}=(1+2i+i^2)^{18}=(2i)^{18}=-2^{18}\\ (2+5i)^2+(2-5i)^2=4+20i+25i^2+4-20i+25i^2=8-25-25=\\ =-42[/tex]
Firemuppets2
[tex] a. (1+i)^{8} = (i^{4}+i)^{8} = (i^{5})^{8} = i^{40} = (i^{4})^{5} = 1 [/tex]
[tex] b. (1+i)^{36} = (i^{4}+i)^{36} = (i^{5})^{36} = i^{180} = (i^{4})^{45} = 1 [/tex]
[tex] c. (2+5i)^{2} + (2-5i)^{2} = 4+20i-25 + 4-20i-25 = 8-50 = -42 [/tex]

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