Let's say that z = a+bi and w = c+di
\( zw - 3w - 2iw + 4z = 7iw + 5w + 11iz - 13 + 15i\)
\(= (a+bi)(c+di) - 3(c+di) - 2i(c+di) + 4(a+bi) = 7i(c+di) + 5(c+di) + 11i(a+bi) - 13 + 15i\)
\(= ac+adi+bci-bd - 3c-3di - 2ic-2di^2 + 4a+4bi = 7ic+7di + 5c+5di + 11ia+11bi^2 - 13 + 15i\)
This is not the correcr way to do it oh gosh...
\(= ac+adi+bci-bd -2di^2 + 4a+4bi = 9ic+15di + 8c + 11ia+11bi^2 - 13 + 15i\)
\( ac-bd+ 4a =+ 8c +(11b^2-4b-bc-ad+2di+9c+15d+11a+15)i - 13\)
yeahhhh..... Im not the one solving this
I want you to do a portion of it to so im going to do half of it.
so first lets simplify this
\(x^2 - 5x + 3 = 2x^2 - 11x + 14\)
oh man... this is crazy..
this is definently not what i had in mind...
ok so now you square them...
I guess you have to find a quadratic...