Question

Find z1z2, when z1=6+2i and z2=2−i.

• A. (A) (1+i)
• B. (B) 2(1+i)
• C. (C) 2+i
• D. (D) (1−i)

Answer 1

Answer: (B)

(B) 2(1+i)

Answer 2

Explanation:
Given,z1=6+2i...(1) z2=2−i...(2)On dividing equation (1) and (2), we getz1z2=6+2i2−iMultiplying by (2+i) in numerator and denominator, we getz1z2=6+2i2−i×2+i2+i⇒z1z2=12+6i+4i+2i222−i2⇒z1z2=12+10i+2×(−1)4−(−1)[∵i2=−1]⇒z1z2=12+10i−24+1⇒z1z2=10+10i5⇒z1z2=10(1+i)5⇒z1z2=2(1+i)Hence, the correct option is (B).