How do you solve #z^2-3iz-(3+i) = 0# ?

2 Answers
May 24, 2017

#z=-1+i" "# or #" "z=1+2i#

Explanation:

Note that:

#abs(3+i) = sqrt(3^2+1^2) = sqrt(10) = sqrt(2)sqrt(5)#

Hence #(3+i)# may be the product of two complex numbers of absolute values #sqrt(2)# and #sqrt(5)#, namely of the forms:

#+-1+-i" "# and #" "+-2+-i" "# or #" "+-1+-2i#

We would like to find two factors of #(3+i)# which differ by #3i#

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#

May 24, 2017

#z = 1+2i" "# or #" "z = -1+i#

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 #abs(sqrt(3+4i)) = sqrt(abs(3+4i)) = sqrt(5)#

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#