Title: Complex Numbers #2
Course: Demonstrating STACK (v4.4.7)

Consider the complex number \[z = [value].\] Give the modulus and argument of all its cube roots.
Use each row in the table below to specify a cube root, with the modulus in the first column and the argument in the second. Each argument should be entered as a multiple of \(\pi\) and should fall in the interval \([-\pi,\pi ]\).
Feedback: For the complex number \[z = [value]\] The modulus \(|z| = [value]\) and the argument of \(z = [value]\) Therefore the cube roots of \(z\) have a modulus of \[\sqrt[3]{z} = \sqrt[3][value] = [value]\] \[\] The arguments of \(z = [value]\) or \([value]\) or \([value]\) Giving the arguments of the cube roots as \( \frac[value]{3}\) ,\(\frac[value]{3}\) and \(\frac[value]{3}\) equal to \([value]\),\([value]\),\([value]\).

Given that \(z\) is the complex number \[[value],\] evaluate the modulus and the principal value of the argument of \([value]\). Give your answers in the correct order i.e. \([r,\theta]\)
Feedback: Let \(r_1\) be the modulus of \(z\) and \(\theta_1\) be the argument. Then \[r_1=\sqrt{x^2+y^2}=\sqrt{\left([value]\right)^2+\left([value]\right)^2}=\sqrt[value]=[value],\] and \[\tan(\theta_1)=\frac[value][value]=[value], \] giving \[\theta_1=[value].\] Therefore for \([value]\)\[r = r_1^[value]=[value]^[value]=[value],\ \] and, \[\theta=[value]\cdot \theta_1=[value]\cdot [value]=[value],\ \] and the principal argument is \[\theta=[value]\]

Given that \(z\) is the complex number \[[value]\] evaluate \([value]\). Give your answers in the form \(p + qi\).
Feedback: Let \(r_1\) be the modulus of \(z\) and \(\theta_1\) be the argument \[ r1=\sqrt{x^2+y^2}=\sqrt{\left([value]\right)^2+\left([value]\right)^2}=\sqrt[value]=[value]\] \[\tan(\theta_1)=\frac[value][value]=[value]\] \[\theta_1=[value]\] Therefore \[r = r_1^[value]=[value]^[value]=[value]\] \[\theta=[value]\cdot \theta_1=[value]\cdot[value]=[value]\] and the principal argument is \[\theta=[value]\] Hence, the real part is found as \[p = [value] \cos([value]) = [value]\cdot [value] = [value],\] and the imaginary part is found as \[q = [value] \sin([value]) = [value]\cdot [value] = [value]\] Hence the answer is \([value]\)

Determine a fifth root of \(z = [value] \) in the form \( p + q i \).

You may use the commands \( \cos() \) and \( \sin() \) in your answers.

\( p + q i =\)
Feedback: For the complex number \[z = [value]\] The modulus \(|z| = [value]\) and the argument of \(z = [value]\) Therefore the cube roots of \(z\) have a modulus of \[\sqrt[3]{z} = \sqrt[3][value] = [value]\] \[\] The arguments of \(z = [value]\) or \([value]\) or \([value]\) Giving the arguments of the cube roots as \( \frac[value]{3}\) ,\(\frac[value]{3}\) and \(\frac[value]{3}\) equal to \([value]\),\([value]\),\([value]\).

Determine a complex root of the cubic equation \[ [value] = 0.\] Hint: the answer should be given in the form \(p + qi\).
Feedback: Firstly, look for a real root by dividing the equation by \([value]\) where \(a\) is a factor of the constant [value]. In this case \(a = [value]\) is the appropriate value. Hence dividing [value] by [value] gives \[[value]\] Using the quadratic formula \[x=\frac{-b±\sqrt{b^2-4ac}}{2a}\] for the equation \(ax^2+bx+c\) gives \[x= \frac{[value]±[value]}[value]\] Therefore \( x = [value],[value]\)

Solve the following quadratic equation: \[[value] = 0\] Hint: give your answer as a set, e.g. {5, 21+3i}. Marks are only given for a complete and correct answer.
The solutions are:
Feedback: Using the quadratic formula \[z=\frac{-b±\sqrt{b^2-4ac}}{2a}\] for the solutions of the equation \(az^2+bz+c=0\) to solve [value] gives \[z= \frac{[value]\pm\sqrt{[value]^2-4\times[value]\times[value]}}[value]=\frac{[value]\pm[value]}[value]=[value]\]

If \(\ [value]\) + \([value] i\ \) is a root of the quartic equation \[[value] = 0,\] find the remaining roots. Hint: give your answer as a set, e.g. {5, 9, 21+3*i}.
The remaining solutions are:
Marks are only given for a complete and correct answer.
Feedback: Given \(\ [value]+[value]i\ \) is a root of the equation \([value] = 0\), then \(\ [value]-[value]i\ \) is a root as well and \[ [value] \] is a factor Dividing by \([value]\) gives \[[value]\] and hence \[[value]=0\]. Therefore the remaining roots are \([value]\)