For 9i − 3, write the conjugate of the complex number, then multiply the number and its conjugate.

Question

darkbluechocobo · Answer

is this -3 + 9i? and the answer is 90?

ganeshie8 · Answer

write given complex number in standard form : a + ib

ganeshie8 · Answer

after that, the congugate would be : a - ib

ganeshie8 · Answer

you just need to flip is sign of imaginary part for conjugate

darkbluechocobo · Answer

ohh so switch it around to -3+9i then it would be -3-9i

ganeshie8 · Answer

yes!

darkbluechocobo · Answer

How would you find the second part the wording confuses me

ganeshie8 · Answer

if you're formula oriented :
$$\large  (a+ib)(a-ib) = a^2 + b^2$$

ganeshie8 · Answer

its same as difference of squares formula :
$$\large   (x+y)(x-y) =x^2 - y^2$$

darkbluechocobo · Answer

I got 90?

ganeshie8 · Answer

$$\large (-3+9i)(-3-9i) =  (-3)^2 + (9)^2$$

ganeshie8 · Answer

90 is $\large \color{red}{\checkmark }$

darkbluechocobo · Answer

whoop so the full answer is -3-9i, 90 then

ganeshie8 · Answer

looks good to me, if its first time you must be excited seeing how multiplication of two complex numbers has produced a pure real number ?

darkbluechocobo · Answer

Oh yes aha. I am quite find of this unit so far. But The only thing im really hating about this at the moment is finding zeros aha