Consider the function with rule \(f(z)=z^4+6 z^2+25\), where \(z \in C\).
- Consider \(z_1=1+2 i\).
- Plot and label \(z_1\) and \(\overline{z}_1\) on the Argand plane below. (1 mark)
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- Given that \(1+2 i\) is a solution of \(f(z)=0\), find a quadratic factor of \(f(z)\). (2 marks)
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- Hence, find all remaining solutions of \(f(z)=0\). (2 marks)
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a. \(z_1=1+z_i \ \Rightarrow \ \overline{z}_1=1-z_i\)
b. \(z^2-2 z+5\)
c. \(z=-1+2 i, z=-1-2 i\)
a. \(z_1=1+z_i \ \Rightarrow \ \overline{z}_1=1-z_i\)
b. \(\text{Since} \ \ 1+2 i \ \ \text{is a solution of}\ \ f(z)=0:\)
\(\Rightarrow 1-2 i \ \ \text{is also a solution (conjugate factor theorem).}\)
\(\text {Express as a quadratic factor:}\)
| \((z-(1+2 i))(z-(1-2 i))\) | \(=((z-1)-2 i)((z-1)+2 i)\) |
| \(=(z-1)^2-4 i^2\) | |
| \(=z^2-2 z+5\) |
c. \(f(z)=z^4+6 z^2+25, \quad z \in C\)
\(\text{Since \(\ z^2-2 z+5\ \) is a factor:}\)
| \(f(z)\) | \(=\left(z^2-2 z+5\right)\left(z^2+a z+b\right)\) |
| \(=z^2\left(z^2+a z+b\right)-2 z\left(z^2+a z+b\right)+5\left(z^2+a z+b\right)\) | |
| \(=z^4+a z^3+b z^2-2 z^3-2 a z^2-2 b z+5 z^2+5 a z+5 b\) | |
| \(=z^4+(a-2) z^3+(b-2 a+5) z^2+(-2 b+5 a) z+5 b\) |
\(\text{Equating co-efficients:}\)
\(a-2=0 \ \Rightarrow \ a=2\)
\(5 b=25 \ \Rightarrow \ b=5\)
\(f(z)=\left(z^2-2 z+5\right)\left(z^2+2 z+5\right)\)
\(\text{Solve:} \ \ z^2+2 z+5=0\)
\(z=\dfrac{-2 \pm \sqrt{2^2-4 \cdot 1 \cdot 5}}{2}=\dfrac{-2 \pm \sqrt{-16}}{2}=-1 \pm 2 i\)
\(\therefore \ \text{Other solutions:}\ \ z=-1+2 i, z=-1-2 i\)
