Question #326636

(1+3i÷2-5i)^2


Expert's answer

(1+3i2−5i)2=(1∗2−5∗1i+2∗3i+3∗529)2==(17−i29)2=288−34i841(\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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Canada #339914 on Dec 2023