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

Hi, can anyone show me how to do this problem? 100 points for this. Thanks in advance-example-1
User PsychOle
by
6.7k points

2 Answers

6 votes

Answer:

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


User Petmez
by
7.0k points
3 votes

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

User Ryan K
by
6.3k points