19M.2.AHL.TZ1.H_2

pestleMathematicsAAHLPaper 219M· ahl-1-14-complex-roots-of-polynomials-conjugate-roots-de-moivres-powers-&-roots-of-complex-numberssource ↗

Solve ${z^2} = 4{{\text{e}}^{\frac{\pi }{2}{\text{i}}}}$ , giving your answers in the form

$r{{\text{e}}^{{\text{i}}\theta }}$ where $r$ , $\theta \in \mathbb{R}$ , $r > 0$ .

[3]

$a + {\text{i}}b$ where $a$ , $b \in \mathbb{R}$ .

[2]

Markscheme / solution

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

$z = 2{{\text{e}}^{\frac{\pi }{4}{\text{i}}}}\,\left( { = 2{{\text{e}}^{0.785{\text{i}}}}} \right)$ A1

Note: Accept all answers in the form $2{{\text{e}}^{\left( {\frac{\pi }{4} + 2\pi n} \right){\text{i}}}}$ .

$z = 2{{\text{e}}^{\frac{{5\pi }}{4}{\text{i}}}}\,\left( { = 2{{\text{e}}^{3.93{\text{i}}}}} \right)$ OR $z = 2{{\text{e}}^{ - \frac{{3\pi }}{4}{\text{i}}}}\,\left( { = 2{{\text{e}}^{ - 2.36{\text{i}}}}} \right)$ (M1)A1

Note: Accept all answers in the form $2{{\text{e}}^{\left( { - \frac{{3\pi }}{4} + 2\pi n} \right){\text{i}}}}$ .

Note: Award M1A0 for correct answers in the incorrect form, eg $- 2{{\text{e}}^{\frac{\pi }{4}{\text{i}}}}$ .

[3 marks]

$z = 1.41 + 1.41{\text{i}}$ , $z = - 1.41 - 1.41{\text{i}}$ A1A1

$\left( {z = \sqrt 2 + \sqrt 2 {\text{i}},\,\,z = - \sqrt 2 - \sqrt 2 {\text{i}}} \right)$

[2 marks]

Examiners’ report

[N/A]