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Exercise 1.2 · Q79

Q.Find the value of x3+x2−x+22x^3+x^2-x+22, if x=51−2ix = \dfrac{5}{1-2i}.

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Simplify x=dfrac51−2i=dfrac5(1+2i)(1−2i)(1+2i)=dfrac5(1+2i)1+4=dfrac5(1+2i)5=1+2ix=\\dfrac{5}{1-2i}=\\dfrac{5(1+2i)}{(1-2i)(1+2i)}=\\dfrac{5(1+2i)}{1+4}=\\dfrac{5(1+2i)}{5}=1+2i. Substitute into x3+x2−x+22x^3+x^2-x+22. x2=(1+2i)2=1+4i+4i2=1+4i−4=−3+4ix^2=(1+2i)^2=1+4i+4i^2=1+4i-4=-3+4i. $x^3=x^2x=(-3+4i)(1+2i)=-3-6i+4i+8i^2=-3-2i-8=-11- …

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