if z is a complex number, find all z that satisfy

z^2 + 2z + 1 =0

Question

if z is a complex number, find all z that satisfy

z^2 + 2z + 1 =0

slaaibak · Answer

$$z = -1 + 0i$$

anonymous · Answer

can you really just factor normally?

slaaibak · Answer

yes, it's (z+1)^2=0

anonymous · Answer

that makes sense to me, thanks

slaaibak · Answer

or use this http://www.purplemath.com/modules/quadform.htm

anonymous · Answer

how about if the equation is z^2 + 2(conjugate z) + 1 = 0

anonymous · Answer

zarkon are you there?

zarkon · Answer

yes

anonymous · Answer

can you help me with this one?
if z is a complex number, find all z that satisfy

z^2 + 2(conjugate z) + 1 = 0

zarkon · Answer

let $z=a+bi$

plug into your equation

set the real and imaginary parts equal to zero...solve for a,b

anonymous · Answer

how can I solve for a and b when I only have 1 equation?

zarkon · Answer

I get $z=-1,1\pm2i$

zarkon · Answer

$z^2+2\overline{z}+1=0$

$(a+bi)^2+2(a-bi)+1=0$

expand...you will and imaginary part and a real part (thus you will have two equations)

solve for a,b

zarkon · Answer

$a^2+2a-b^2+1+(2ab-2b)i=0$

thus 

$a^2+2a-b^2+1=0$ and $2ab-2b=0$

anonymous · Answer

so in order to have that complex thing be 0 either the real or the imaginary part has to be 0

which means 2 equations that you can solve. Brilliant!

zarkon · Answer

then both have to be zero

a+bi=0

same as a+bi=0+0i

equate real and imaginary parts

anonymous · Answer

right woops

anonymous · Answer

that makes a lot of sense though. Thanks for your help!

zarkon · Answer

no problem