Complex Numbers: a=1+4i satisfies z^3+5z^2+kz+m=0, k,m real constants

TToby

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Jun 12, 2018
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So I have a = 1 + 4i that satisfies the cubic equation:

z3+5z2+kz+m=0 where k and m are real and constant

How can I work out k and m?

Thank you for your help
 

Dr.Peterson

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Nov 12, 2017
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So I have a = 1 + 4i that satisfies the cubic equation:

z3+5z2+kz+m=0 where k and m are real and constant

How can I work out k and m?

Thank you for your help
Just plug in 1 + 4i for z, expand, and set the real and imaginary parts equal to zero. You will then have two equations to solve for two unknowns.

(1 + 4i)3+5(1 + 4i)2+k(1 + 4i)+m=0

If you had shown some work or told us where you need help, I could have said more; as it is, this should get you started. If you get stuck, show us where.
 

ksdhart2

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Mar 25, 2016
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If a value "satisfies a cubic equation" means that when you plug that value to the cubic, you get a true equation. So let's do that now.

\(\displaystyle (1 + 4i)^3 + 5(1+ 4i)^2 + k(1 + 4i) + m = 0\)

This should be easy enough for you to solve for m in terms of k, arriving at:

\(\displaystyle m = \text {(Real part as a function of k)} + i \text {(Imaginary part as a function of k)}\)

The problem text tells you that both m and k must be real, so what can you conclude about the imaginary part of m? What does that tell you about the value of k? What does that, then, tell you about the value of m?
 
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