How do you simplify $(7 - 6 i) (- 8 + 3 i)$ $(7-6i)(-8+3i)$ ?

Question

6641 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-09-02T14:43:10

$(7 - 6 i) (- 8 + 3 i) = - 38 + 69 i$ $(7-6i)(-8+3i)=-38+69i$

While doing any multiplication or division exercise involving complex numbers, one must always remember that $i^{2} = - 1$ $i^2=-1$ .

Hence, $(7 - 6 i) (- 8 + 3 i)$ $(7-6i)(-8+3i)$

= $7 \times (- 8) + 7 \times 3 i - 6 i \times (- 8) - 6 i \times 3 i$ $7×(-8)+7×3i-6i×(-8)-6i×3i$

= $- 56 + 21 i + 48 i - 18 i^{2}$ $-56+21i+48i-18i^2$

= $- 56 + 69 i - 18 \times (- 1)$ $-56+69i-18×(-1)$

= $- 56 + 69 i + 18$ $-56+69i+18$

= $- 38 + 69 i$ $-38+69i$

How do you simplify (7-6i)(-8+3i)(7−6i)(−8+3i)(7-6i)(-8+3i)?