Question

Find the product of the complex number and its conjugate.
1 + 3i

Answer 1

so, what's its conjugate anyway?

Answer 2

would it be 1-3i?

Answer 3

yeap, so the product, will be \(\bf (1+3i)(1-3i)\\
\textit{keep in mind that } (a^2-b^2) = (a-b)(a+b)\)

Answer 4

would 1-9i be correct? or is there more to it then that?

Answer 5

... what's \(\bf \large i\) equals to again?

Answer 6

square root of -1

Answer 7

I see... so \(\bf (1+3i)(1-3i)\\
\textit{keep in mind that } (a^2-b^2) = (a-b)(a+b)\\
(1+3i)(1-3i) \implies 1^2-(3i)^2 \implies 1-3^2i^2 \\\quad \\
\implies 1-(3\times3)(\sqrt{-1}\times \sqrt{-1}) \implies 1-(9)(\sqrt{(-1)^2})\)

what would that give you?

Answer 8

\(\large 1-(9)\left(\sqrt{(-1)^2}\right)\) what do you think?

Answer 9

10

Answer 10

\(\bf 1-(9)\left(\sqrt{(-1)^2}\right) \implies
1-9(-1) \implies 10\)

Answer 11

yeap

Answer 12

thank you!

Answer 13

yw

Answer 14

(1+3i)(1-3i)=1-9(i^2)=1+9=10
conj=1-3i