Question

Multiply. Write the result in the form a+bi

3i(2-3i)

(4+5i)(2+i)

(-1+6i)(3-2i)

Answer 1

Number 1: 3i(2-3i)

`\mathrm{Apply\:complex\:arithmetic\:rule}:\quad \left(ai\right)\left(b+ci\right)=-ac+abi\\a=3,\:b=2,\:c=-3\\=-3\left(-3\right)+3\cdot \:2i\\Refine\\=9+6i\\`

Number 2: (4+5i)(2+i)

`\mathrm{Apply\:complex\:arithmetic\:rule}:\quad \left(a+bi\right)\left(c+di\right)=\left(ac-bd\right)+\left(ad+bc\right)i\\a=4,\:b=5,\:c=2,\:d=1\\=\left(4\cdot \:2-5\cdot \:1\right)+\left(4\cdot \:1+5\cdot \:2\right)i\\Refine\\=3+14i`

Number 3: (-1+6i)(3-2i)

`\mathrm{Apply\:complex\:arithmetic\:rule}:\quad \left(a+bi\right)\left(c+di\right)=\left(ac-bd\right)+\left(ad+bc\right)i\\a=-1,\:b=6,\:c=3,\:d=-2\\=\left(-1\cdot \:3-6\left(-2\right)\right)+\left(-1\cdot \left(-2\right)+6\cdot \:3\right)i\\Refine\\=9+20i`