Finding the Value of m in a Complex Number Equation

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Saitama
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Homework Statement


Let z be a complex number satisfying the equation ##z^3-(3+i)z+m+2i=0##, where mεR. Suppose the equation has a real root, then find the value of m.

Homework Equations


The Attempt at a Solution


The equation has one real root which means that the other two roots are complex and are conjugates of each other.
Let ##\alpha, \beta, \gamma## be the three roots. The coefficient of z^2 is zero.
Hence ##\alpha+\beta+\gamma=0##. Let ##\gamma## be the real root and the other two are complex. The sum of the complex roots is zero. From here, i get ##\gamma=0##.
Product of roots, ##\alpha \beta \gamma=m+2i=0##. This gives me ##m=-2i## which is incorrect as m is a real number.

Any help is appreciated. Thanks!
 
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The sum of the complex roots is zero.
Why? Their imaginary parts cancel, but the real parts do not have to.

Your approach is way too complicated. If the equation has a real root, there is a real z which satisfies the equation. What is the imaginary part of the left side? It has to be zero...
This allows to calculate z and m.
 
mfb said:
Why? Their imaginary parts cancel, but the real parts do not have to.

Oops, sorry about that.

mfb said:
Your approach is way too complicated. If the equation has a real root, there is a real z which satisfies the equation. What is the imaginary part of the left side? It has to be zero...
This allows to calculate z and m.
Do you mean I have to substitute z=x+iy and compare the real and imaginary parts?
 
mfb said:
there is a real z which satisfies the equation.
What is the imaginary part of z^3 if z is real? What is the imaginary part of (3+i)z?
m is real, and the imaginary part of 2i is obvious.
 
Thanks a lot mfb! :smile: