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Worked Examples · Example 3

Q.Simplify (3+2i)(1−4i)(3+2i)(1-4i) and write the result in the form a+bia+bi.

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✓ Free question

Expand every term:

(3+2i)(1−4i)=3(1)+3(−4i)+2i(1)+2i(−4i)=3−12i+2i−8i2.(3+2i)(1-4i)=3(1)+3(-4i)+2i(1)+2i(-4i)=3-12i+2i-8i^{2}.

Replace i2=−1i^{2}=-1, so −8i2=+8-8i^{2}=+8:

=3+8+(−12+2)i=11−10i.=3+8+(-12+2)i=11-10i.

Check (independent recomputation): grouping as (real)=3⋅1−2⋅(−4)=3+8=11=3\cdot1-2\cdot(-4)=3+8=11 and (imaginary)=3⋅(−4)+2⋅1=−12+2=−10=3\cdot(-4)+2\cdot1=-12+2=-10, using (a+bi)(c+di)=(ac−bd)+(ad+bc)i(a+bi)(c+di)=(ac-bd)+(ad+bc)i with a=3,b=2,c=1,d=−4a=3,b=2,c=1,d=-4 — gives 11−10i11-10i, matching.

✓Final answer

(3+2i)(1−4i)=11−10i(3+2i)(1-4i)=11-10i

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