Question 853804
<pre>
Z² + (1+i)Z + (5-i4) = 0

That's a quadratic in Z.  Use the quadratic formula:

{{{Z = (-(1+i) +- sqrt((1+i)^2-4(1)(5-i4)))/(2(1)) }}}

{{{Z = (-1-i +- sqrt((1+2i+i^2)-4(5-i4)))/2 }}}

{{{Z = (-1-i +- sqrt(1+2i+i^2-20+16i)))/2 }}}

{{{Z = (-1-i +- sqrt(1+18i+i^2-20)))/2 }}}


Replace i² by (-1)

{{{Z = (-1-i +- sqrt(1+18i+(-1)-20)))/2 }}}

{{{Z = (-1-i +- sqrt(1+18i-1-20)))/2 }}}

{{{Z = (-1-i +- sqrt(-20+18i))/2 = expr(1/2)*(-1-i +- sqrt(-20+18i))}}}

Now we have to find the square root by DeMoivre's theorem:

{{{matrix(2,1,"",-20+18i)^(1/2))}}}

Put -20+18i in polar form

{{{r=sqrt((-20)^2+18^2)=sqrt(400+324)=sqrt(724)}}}

{{{theta=tan^"-1"(18/(-20))=tan^"-1"(-0.9)}}}

{{{-20+18i = sqrt(724)*cis(tan^"-1"(-0.9))}}}

{{{matrix(2,1,"",(-20+18i))^(1/2)=(sqrt(724))^(1/2)cis(expr(tan^"-1"(-0.9))/2))}}}{{{""=""}}}{{{matrix(2,1,"",724^(1/4)cis(expr(tan^"-1"(-0.9))/2))=root(4,724)cis(expr(tan^"-1"(-0.9))/2))}}}



root(4,724)cis(expr(tan^"-1"(-0.9))/2)


{{{Z = expr(1/2)*(-1-i +- root(4,724)cis(expr(tan^"-1"(-0.9))/2))}}}

or decimal approximation (and you can get it immediately with a TI-84
straight from the original equation):

Z = .4291964334+1.92144709i,  -1.429196433-2.92144709i

Edwin</pre>