Question #326636

(1+3i÷2-5i)^2


1
Expert's answer
2022-04-11T15:47:26-0400

(1+3i25i)2=(1251i+23i+3529)2==(17i29)2=28834i841(\cfrac{1+3i}{2-5i})^2 = (\cfrac{1*2 -5*1i+2*3i +3*5}{29})^2 =\\ = (\cfrac{17-i}{29})^2 = \cfrac{288-34i}{841}


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