Solve:z^2-8z+19+4i=0?? thanks in advanced

1 Answer
Feb 24, 2018

#z_1=5-2i# and #z_2=3+2i#

Explanation:

#z^2-8z+19+4i=0#

#Delta=(-8)^2-4*1*(19+4i)=-12-16i=(a+bi)^2#

#(a+bi)^2=a^2-b^2+2abi=-12-16i#

After equating real and imaginary parts of complex number,

#a^2-b^2=-12# and #2ab=-16#

After solving these equations, possible pairs of #(a, b)# are #(-2, 4)# and #(2, -4)#

After choosing #Delta=2-4i#,

#z_1=(8+(2-4i))/2=5-2i# and #z_2=(8-(2-4i))/2=3+2i#