Question

Hi, can anyone show me how to do this problem? 100 points for this.
Thanks in advance

Answer 1

z^2 + (-1 + 5·i)·z + 14 - 7·i = 0

(1 + 2·i)^2 + (-1 + 5·i)·(1 + 2·i) + 14 - 7·i = 0

(1 + 4·i - 4) + (-1 - 2·i + 5·i - 10) + 14 - 7·i = 0

0 = 0 --> true

z^2 + (-1 + 5·i)·z + 14 - 7·i = 0

(1 - 2·i)^2 + (-1 + 5·i)·(1 - 2·i) + 14 - 7·i = 0

(1 - 4·i - 4) + (-1 + 2·i + 5·i + 10) + 14 - 7·i = 0

20 - 4·i = 0 --> false

Answer 2

z^2 + (-1 + 5·i)·z + 14 - 7·i = 0

1+2i is a root

z= 1+2i

z^2 = (1+2i) (1+2i)

= 1 +2i+2i +4i^2

= 1 +4i -4

= -3+4i

= (-1+5i) (1+2i)

-1+5i-2i+10i^2

-1+3i-10

-11+3i

z^2 + (-1 + 5·i)·z + 14 - 7·i = 0

-3+4i + -11+3i +14 - 7i

Combine like terms

-3-11 +14 +4i +3i-7i

0

So 1+2i is a root

1-2i is not a root

z= 1-2i

z^2 = (1-2i) (1-2i)

= 1 -2i-2i +4i^2

= 1 -4i -4

= -3-4i

= (-1+5i) (1-2i)

-1+5i+2i-10i^2

-1+7i+10

9+7i

z^2 + (-1 + 5·i)·z + 14 - 7·i = 0

-3-4i + 9+7i +14 - 7i

Combine like terms

-3+9 +14 -4i +7i-7i

20 -4i

So 1-2i is not a root

The complex conjugate being roots is only true for real coefficients