Find the square roots of the complex number [-80-18i ] and solve the following quadratic equation. [4z^2+(16i-4)z+(65+10i) = 0] Solution Let [sqrt(-80-18i)=a+ib] squaring both side [-80-18i=a^2-b^2+2abi] coparing real and magnary parts both side [a^2-b^2=-80] [2ab=-18] But [(a^2+b^2)^2=(a^2-b^2)^2+4a^2b^2] [(a^2+b^2)=(-80)^2+(-18)^2] [=6400+324=6724] [a^2+b^2=+-82] Now solve [a^2-b^2=-80] [a^2+b^2=+-82] [If] [a^2-b^2=-80] [a^2+b^2=82] [Then] [a=+-1 and b=+-9] [if] [a^2-b^2=-80] [a^2+b^2=-82] [a=+-9 and b=+-1] [Thus] [sqrt(-80-18i)=+-1+-9i] [or] [sqrt(-80-18i)=+-9+-i] [4z^2+(16i-4)z+(65+10i)=0] [z=(-(16i-4)+-sqrt((16i-4)^2-16(65+10i)))/8] [=(-(16i-4)+-sqrt(-256+16-128i-1040-160i))/8(] [=(-(16i-4)+-sqrt(-1280-288i))/8] [=(-(4i-1)+-sqrt(-80-18i))/2] [Thus] [z=(-4i+1+-1+-9i)/2] [and] [z=(-4i+1+-9+-i)/2].