Title: Complex Numbers #1
Course: Demonstrating STACK (v4.4.7)

Consider the complex number \[z=[value] \] Evaluate the modulus of \(z\) \((r = |z|)\) and the principal value of the argument of \(z\) (\(\theta = \mathrm{Arg}(z)\)). Enter your results in the correct order, i.e. \([r,\theta]\).

\([r,\theta] = \)
Feedback: \[r=\sqrt{x^2+y^2}\] \[r=\sqrt{\left([value]\right)^2+\left([value]\right)^2}=\sqrt{[value]+[value]}\] \[r=\sqrt[value]=[value]\] \[tan(\mathrm{arg}(z))=\frac{y}{x}=\frac[value][value]=[value]\] \[\mathrm{arg}(z)=[value] + k\cdot\pi\] Considering the appropriate quadrant of the complex plane to fix \(k\), we find that \[\mathrm{arg}(z)=[value]\].

Consider the complex number \(z\) with modulus \(|z| = [value]\) and argument \(\arg(z) = [value] \).

Convert this number into the form \(p + qi\) where \(p\) and \(q\) are real.
Feedback: To convert the complex number \(z\) with modulus \(|z| = [value]\) and argument \(\arg(z) = [value] \) into the form \(p + qi\) where \(p\) and \(q\) are real we use trigonometry. \[p=r \cos(\theta)=[value]\cdot \cos([value])=[value]\ ,\] and, \[q=r \sin(\theta)=[value]\cdot \sin([value])=[value]\] Hence the answer is \([value]\)

Convert the following number into the form \(p+qi\) where \(p\) and \(q\) are real. Note: give your answer in the form \(p+qi\), in the box below. \[ \frac{[value] + [value]i}{[value]+[value]i}\]
Feedback: Multiplying the top and bottom by the conjugate of the denominator gives \[ \frac{\ [value]+[value]i\ }{\ [value]+[value]i\ }*\frac{\ [value]-[value]i\ }{\ [value]-[value]i\ }= \frac{\ [value]+[value]i-[value]i-[value]i^2\ }{\ [value]+[value]i-[value]i-[value]i^2\ }\] and because \[i^2= -1\] \[ \frac{\ [value]+[value]i-[value]i-[value]i^2\ }{\ [value]+[value]i-[value]i-[value]i^2\ }=\frac{\ [value]\ }{\ [value]\ }=[value]\]

Convert the following number into the form \(p+qi\) where \(p\) and \(q\) are real. Note: give your answer in the form \(p+qi\), in the box below. \[ \frac{2}[value] + \frac{5}{ [value] } \]
Feedback: Multiplying the top and bottom by the conjugate of the denominator gives \[ \frac{2}[value]*\frac[value][value]+\frac{5}[value]*\frac[value][value]=\] \[\frac{[value]-[value]i}{[value]-[value]i+[value]i-[value]i^2}+\frac{[value]+[value]i}{[value]-[value]i+[value]i-[value]i^2} \] and because \[i^2= -1\] becomes \[\frac{[value]-[value]i}[value]+\frac{[value]+[value]i}[value]=\] \[\frac{[value]-[value]i}[value]+\frac{[value]+[value]i}[value]=[value]\]

Find, in the form \(a + bi\), where \(a\) and \(b\) are real, a square root of \[[value]\]
\(a + bi\) =
Feedback: Let \[\sqrt[value] = a+ib\] Squaring gives us \[[value] = a^2 + 2abi - b^2\] Therefore equating the real part\[[value] = a^2-b^2\] and the imaginary part\[[value] = 2ab\] Hence \[b=\frac[value]{2a}=\frac[value]{a}\] \[[value]=[value]-\frac[value][value]\] \[[value]=0\] Finding the positive solution of this equation as a quadratic in [value] gives \[[value]=\frac{[value]+\sqrt{[value]+[value]}}{2}=\frac{[value]+\sqrt[value]}{2}=\frac{[value]+[value]}{2}=[value]\] Therefore \[ a= [value] \mbox{ and } b = \frac[value][value]=[value]\] or \[ a= [value] \mbox{ and } b = \frac[value][value]=[value]\]