How do you solve #z^2-3iz-(3+i) = 0# ?
2 Answers
Explanation:
Note that:
#abs(3+i) = sqrt(3^2+1^2) = sqrt(10) = sqrt(2)sqrt(5)#
Hence
#+-1+-i" "# and#" "+-2+-i" "# or#" "+-1+-2i#
We would like to find two factors of
We find:
#(1-i)(1+2i) = 3+i#
Hence:
#0 = z^2-3iz-(3+i) = (z+1-i)(z-1-2i)#
So:
#z=-1+i" "# or#" "z=1+2i#
Explanation:
Given:
#z^2-3iz-(3+i) = 0#
The quadratic formula gives us:
#z = (3i+-sqrt((-3i)^2-4(1)(-(3+i))))/(2*1)#
#color(white)(z) = (3i+-sqrt(-9+12+4i))/2#
#color(white)(z) = (3i+-sqrt(3+4i))/2#
Note that:
#abs(3+4i) = sqrt(3^2+4^2) = sqrt(9+16) = sqrt(25) = 5#
So
We find:
#(2+i)^2 = 4+4i+i^2 = 3+4i#
So:
#sqrt(3+4i) = 2+i#
So the roots of the given equation are:
#z = (3i+-sqrt(3+4i))/2#
#z = (3i+-(2+i))/2#
That is:
#z = (3i+(2+i))/2 = 1+2i#
and:
#z = (3i-(2+i))/2 = -1+i#