Hi.
Given: \(z=a+bi\)
So: \(z^3=(a+bi)^3=a^3+3a^2bi-3ab^2-b^3i\)
Given: \(z^3+\left | z\right |^2+z=0\)
Substitute \(z,z^3\), and we know: \(\left |z \right |^2=\sqrt{a^2+b^2}^2=a^2+b^2\)
\(a^3+3a^2bi-3ab^2-b^3i+(a^2+b^2)+a+bi=0\)
We equate real parts = 0 and imaginary parts = 0 (As the right-hand side is zero, this means both the real and imaginary parts are zero.) (Real parts without the "i" and imaginary parts with the "i").
So: Set \(Re=0:\) \(a^3-3ab^2+a^2+b^2+a=0\) (1)
Set \(Im=0:\) \(3a^2b-b^3+b=0\) (2)
Given b is not equal to zero so, divide (2) by b:
\(3a^2-b^2+1=0\)
Let's make b the subject of the equation: (That is, solve for b.)
\(b=\sqrt{3a^2+1}\)
Hence, substituting b in (1) to get:
\(a^3-3a(3a^2+1)+a^2+3a^2+1+a=0\) (Expand.)
\(a^3-9a^3-3a+a^2+3a^2+1+a=0\) (Simplify.)
\(-8a^3-2a+4a^2+1=0\) (Multiply by -1)
\(8a^3-4a^2+2a-1=0\)
We can factor this by grouping:
\(4a^2(2a-1)+(2a-1)=0\)
\((2a-1)(4a^2+1)=0\)
We know that a is a positive real number.
So we only take: \(a=\frac{1}{2}\)
Hence, \(b=\sqrt{3(0.5)^2+1}=\sqrt{\frac{7}{4}}=\frac{\sqrt{7}}{2}\)
Therefore, \((\frac{1}{2},\frac{{\sqrt{7}}}{2})\) is the desired answer.
I hope this helps :)!
