Solved: Consider the equation $(z-1)^3=i$, $z \in \mathbb{C}$. The roots of this... [IB Mathematics Analysis and Approaches (AA)]

Question

HLPaper 1

Consider the equation $(z-1)^3=i$ , $z \in \mathbb{C}$ . The roots of this equation are $\omega_1$ , $\omega_2$ and $\omega_3$ , where $\Im\left(\omega_2\right)>0$ and $\Im\left(\omega_3\right)<0$ . The roots $\omega_1$ , $\omega_2$ and $\omega_3$ are represented by the points $X$ , $Y$ and $Z$ respectively on an Argand diagram. Consider the equation $(z-1)^3=i(z^3)$ , $z \in \mathbb{C}$ .

Verify that ${\omega}_1=1+e^{i\frac{\pi}{6}}$ is a root of this equation.

[2]

Verified

Solution

${1 + e^{i\frac{\pi}{6}} - 1^3}$ ${= e^{i\frac{\pi}{6}}^3}$ A1 ${= e^{i\frac{\pi}{2}}}$ A1 ${= \cos\left(\frac{\pi}{2}\right) + i\sin\left(\frac{\pi}{2}\right)}$ ${= i}$ AG

Note: Candidates who solve the equation correctly can be awarded the above two marks. The working for part (i) may be seen in part (ii).

[2marks]

Find ${\omega}_2$ and ${\omega}_3$ , expressing these in the form ${a}+{e}^{i\theta}$ , where ${a} \in \mathbb{R}$ and ${\theta} > 0$ .

[4]

Verified

Solution

${(zi-1)^3=e^{i(\frac{\pi}{2}+2\pi k)}}$ (M1) ${z-1=e^{i(\frac{\pi}{6}+\frac{4\pi k}{6})}}$ (M1) ${(k=1)⇒ω_2=1+e^{i(\frac{5\pi}{6})}}$ A1 ${(k=2)⇒ω_3=1+e^{i(\frac{9\pi}{6})}}$ A1

[4 marks]

Plot the points $X$ , $Y$ and $Z$ on an Argand diagram.

[4]

Verified

Solution

EITHER attempt to express $e^{i\frac{\pi}{6}}$ , $e^{i\frac{5\pi}{6}}$ , $e^{i\frac{9\pi}{6}}$ in Cartesian form and translate 1 unit in the positive direction of the real axis (M1) ** OR** attempt to express $w_1$ , $w_2$ and $w_3$ in Cartesian form (M1)

THEN Note: To award A marks, it is not necessary to see $X$ , $Y$ or $Z$ , the $w_1$ , or the solid lines A1A1A1

[4marks]

Find $XZ$ .

[3]

Verified

Solution

valid attempt to find ${\omega}_1 - {\omega}_3$ or ${\omega}_3 - {\omega}_1$ M1 ${\omega}_1 - {\omega}_3 = (1 + \frac{{\sqrt{3}}}{2} + \frac{1}{2}i) - (1 - i) = \frac{{\sqrt{3}}}{2} + \frac{3}{2}i$ OR ${\cos}(\frac{\pi}{6}) + i \; {\sin}(\frac{\pi}{6}) + i \; {\sin}(\frac{\pi}{2})$ valid attempt to find ${|\frac{{\sqrt{3}}}{2} + \frac{3}{2}i|}$ M1 ${{\sqrt{\frac{3}{4}} + \frac{9}{4}}}$ ${XZ} = {\sqrt{3}}$ A1

[3marks]*

By using de Moivre’s theorem, show that ${\beta} = \frac{1}{{1 - e^{i\frac{\pi}{6}}}}$ is a root of this equation.

[3]

Verified

Solution

METHOD 1 $\frac{1}{{1 - e^{i\frac{\pi}{6}}}} = \frac{1}{{1 - (\cos\left(\frac{\pi}{6}\right) + i\sin\left(\frac{\pi}{6}\right))}} \quad \emph{M1}$ $= \frac{2}{{2 - \sqrt{3} - i}} \quad \emph{A1}$ attempt to use conjugate to rationalise \quad \emph{M1} $= \frac{{4 - 2\sqrt{3} + 2i}}{{(2 - \sqrt{3})^2 + 1}} \quad \emph{A1}$ $= \frac{1}{2} + \frac{1}{{4 - 2\sqrt{3}}}i$ $\Rightarrow Re(\beta) = \frac{1}{2} \quad \emph{A1}$

Note: Their final imaginary part does not have to be correct in order for the final three $A$ marks to be awarded

METHOD 2 $\frac{1}{{1 - e^{i\frac{\pi}{6}}}} = \frac{1}{{1 - (\cos\left(\frac{\pi}{6}\right) + i\sin\left(\frac{\pi}{6}\right))}} \quad \emph{M1}$ attempt to use conjugate to rationalise \quad \emph{M1} $= \frac{1}{{(1 - \cos\left(\frac{\pi}{6}\right)) - i\sin\left(\frac{\pi}{6}\right)}} \times \frac{{(1 - \cos\left(\frac{\pi}{6}\right)) + i\sin\left(\frac{\pi}{6}\right)}}{{(1 - \cos\left(\frac{\pi}{6}\right)) + i\sin\left(\frac{\pi}{6}\right)}} \quad \emph{A1}$ $= \frac{{(1 - \cos\left(\frac{\pi}{6}\right)) + i\sin\left(\frac{\pi}{6}\right)}}{{(2) - 2\cos\left(\frac{\pi}{6}\right)}} \quad \emph{A1}$ $= \frac{{(1 - \cos\left(\frac{\pi}{6}\right)) + i\sin\left(\frac{\pi}{6}\right)}}{{2 - 2\cos\left(\frac{\pi}{6}\right)}} \quad \emph{A1}$ $= \frac{1}{2} + \frac{1}{{2 - 2\cos\left(\frac{\pi}{6}\right)}}i$ $\Rightarrow Re(\beta) = \frac{1}{2} \quad \emph{A1}$

Note: Their final imaginary part does not have to be correct in order for the final three $A$ marks to be awarded

METHOD 3 attempt to multiply through by $\left(-\frac{e^{-i\frac{\pi}{12}}}{e^{-i\frac{\pi}{12}}}\right) \quad \emph{M1}$ $\frac{1}{{1 - e^{i\frac{\pi}{6}}}} = -\frac{{e^{-i\frac{\pi}{12}}}}{{e^{-i\frac{\pi}{12}} - e^{i\frac{\pi}{12}}}} \quad \emph{A1}$ attempting to re-write in r-cis form \quad \emph{M1} $= -\frac{{\cos\left(\frac{-\pi}{12}\right) + i\sin\left(\frac{-\pi}{12}\right)}}{{\cos\left(\frac{\pi}{12}\right) + i\sin\left(\frac{\pi}{12}\right) - \cos\left(\frac{-\pi}{12}\right) - i\sin\left(\frac{-\pi}{12}\right)}} \quad \emph{A1}$ $= -\frac{{\cos\left(\frac{-\pi}{12}\right) + i\sin\left(\frac{-\pi}{12}\right)}}{{2i\sin\left(\frac{\pi}{12}\right)}} \quad \emph{A1}$ $= \frac{1}{2} - \frac{1}{2i}\cot\left(\frac{\pi}{12}\right)$ $\Rightarrow Re(\beta) = \frac{1}{2} \quad \emph{A1}$

METHOD 4 attempt to multiply through by $\left(\frac{{1 - e^{-i\frac{\pi}{6}}}}{{1 - e^{-i\frac{\pi}{6}}}}\right) \quad \emph{M1}$ $\frac{1}{{1 - e^{i\frac{\pi}{6}}}} = \frac{{1 - e^{-i\frac{\pi}{6}}}}{{1 - e^{-i\frac{\pi}{6}} - e^{i\frac{\pi}{6}} + 1}} \quad \emph{A1}$ attempting to re-write in r-cis form \quad \emph{M1} $= \frac{{1 - \cos\left(\frac{\pi}{6}\right) - i\sin\left(\frac{\pi}{6}\right)}}{{(2 - \cos\left(\frac{\pi}{6}\right))}} \quad \emph{A1}$ attempt to re-write in Cartesian form \quad \emph{M1} $= \frac{{1 - \frac{\sqrt{3}}{2} - \frac{1}{2}i}}{{2 - \sqrt{3}}} \quad \emph{A1}$ $\Rightarrow Re(\beta) = \frac{1}{2} \quad \emph{A1}$

Note: Their final imaginary part does not have to be correct in order for the final $A$ mark to be awarded

[6 marks]

Consider the equation $$(z-1)^3=i$$, $$z \in \mathbb{C}$$. The roots of this equation are $$\omega_1$$, $$\omega_2$$ and $$\omega_3$$, where $$\Im\left(\omega_2\right)>0$$ and $$\Im\left(\omega_3\right)<0$$.

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