Question

find the product z1*z2 and quotient z1/z2 in 2 ways using the standard form for z1 and z2. z1=2-3i, z2=1-sqrt3i

Answer 1

\[(a+bi)(c+di)=(ac-bd)+(ad+bc)i\]

Answer 2

\[(2-3i)(1-\sqrt{3}i)=(2\times 1-3\times \sqrt{3})+(-2\sqrt{3}-3)i\]
is a start

Answer 3

ok

Answer 4

how do i do the dividion part?

Answer 5

do divide, multiply top and bottom by the conjugate of the bottom. the conjugate of \(a+bi\) is \(a-bi\) and this works because \((a+bi)(a-bi)=a^2+b^2\) a real number

Answer 6

\[\frac{2-3i}{1-\sqrt{3}i}=\frac{2-3i}{1-\sqrt{3}i}\times \frac{1+\sqrt{3}i}{1+\sqrt{3}i}\]
\[=\frac{(2-3i)(1-\sqrt{3}i)}{1+3}\] etc

Answer 7

now your only real job is to multiply out in the numerator