"Find the real number k for which 1 + ki, (i = √-1), is a zero of the polynomial z² + kz + 5"

I get k=3 or k=-2, but according to the answer sheet k=-2 is the only answer. Can someone explain why?

Question

"Find the real number k for which 1 + ki, (i = √-1), is a zero of the polynomial z² + kz + 5"

I get k=3 or k=-2, but according to the answer sheet k=-2 is the only answer. Can someone explain why?

anonymous · Answer

Yes, and I get k= 3 Or k=-2

anonymous · Answer

it says k is real

anonymous · Answer

However only k=-2 is supposed to be correct

anonymous · Answer

z² + kz + 5 

roots : 1 + ki, and 1 - ki

anonymous · Answer

sum of roots :
2 = -k

anonymous · Answer

product of roots :
1 + k^2 = 5
k^2 = 4
k = +2, - 2

anonymous · Answer

so i am getting k = 2 and -2

anonymous · Answer

Wouldn't (1+ki) + (1-ki) be 2?

anonymous · Answer

stgreen can you show your working out? Because I really don't know what to do here :P

whpalmer4 · Answer

$$(1+ki)^2+k(1+ki)+5=0$$
$$(1+2ki-k^2)+k+k^2i+5=0$$

To get rid of terms in $i$ $$2ki=-k^2$$

whpalmer4 · Answer

Uh should be $$2ki=-k^2i$$

whpalmer4 · Answer

Then $k=-2$

anonymous · Answer

I get everything except how you made the jump from (1+2ki−k2)+k+k2i+5=0 to 2ki=−k2i. Can you explain that part?

anonymous · Answer

Aaa right I get it now, thanks a bunch!

whpalmer4 · Answer

Great! Glad you were able to figure it out...it's a handy trick!  Of course you have to verify that everything else sums to 0, but if that didn't, you wouldn't have a solution.