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Miscellaneous Exercise 1 (Subjective) · Q176

Q.Find the value of x3+2x2−3x+21x^3+2x^2-3x+21, if x=1+2ix = 1+2i.

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x=1+2ix=1+2i. 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. x3=x2x=(−3+4i)(1+2i)=−3−6i+4i+8i2=−3−2i−8=−11−2ix^3=x^2x=(-3+4i)(1+2i)=-3-6i+4i+8i^2=-3-2i-8=-11-2i. So $x^3+2x^2-3x+21=(-11-2i)+2(-3+4i)-3(1+2i)+21=(-11-2i) …

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