Answers ∣ z 3 ∣ = 3 . {|z_3| = \sqrt{3}.} ∣ z 3 ∣ = 3 . arg ( z 3 ) = 7 30 π . {\arg (z_3) = \frac{7}{30} \pi.} arg ( z 3 ) = 30 7 π . ∣ z 3 n ∣ = 2187 3 . {\left| z_3^n \right| = 2187 \sqrt{3}.} ∣ z 3 n ∣ = 2187 3 . arg ( z 3 n ) = − 1 2 π . {\arg \left( z_3^n \right) = - \frac{1}{2} \pi.} arg ( z 3 n ) = − 2 1 π . Full solutions
(a) ∣ z 1 ∣ = 3 2 + ( 3 ) 2 = 2 3 \begin{align*}
|z_1| &= \sqrt{3^2 + (\sqrt{3})^2}
\\ &= 2 \sqrt{3}
\end{align*} ∣ z 1 ∣ = 3 2 + ( 3 ) 2 = 2 3 arg ( z 1 ) = − tan − 1 3 3 = − 1 6 π \begin{align*}
\arg (z_1) &= - \tan^{-1} \frac{\sqrt{3}}{3}
\\ &= - \frac{1}{6} \pi
\end{align*} arg ( z 1 ) = − tan − 1 3 3 = − 6 1 π z 3 = z 1 × z 2 = 2 3 e − 1 6 π i × 1 2 e 2 5 π i = 3 e 7 30 π i \begin{align*}
z_3 &= z_1 \times z_2
\\ &= 2 \sqrt{3} \mathrm{e}^{- \frac{1}{6} \pi \mathrm{i}} \times \frac{1}{2} \mathrm{e}^{\frac{2}{5} \pi \mathrm{i}}
\\ &= \sqrt{3} \mathrm{e}^{\frac{7}{30} \pi \mathrm{i}}
\end{align*} z 3 = z 1 × z 2 = 2 3 e − 6 1 π i × 2 1 e 5 2 π i = 3 e 30 7 π i ∣ z 3 ∣ = 3 ■ arg ( z 3 ) = 7 30 π ■ \begin{align*}
|z_3| &= \sqrt{3} \; \blacksquare
\\ \arg (z_3) &= \frac{7}{30} \pi \; \blacksquare
\end{align*} ∣ z 3 ∣ arg ( z 3 ) = 3 ■ = 30 7 π ■
(b)
(c) For
z 3 n {z_3^n} z 3 n to be purely imaginary,
for
m ∈ Z , {m \in \mathbb{Z},} m ∈ Z , arg ( z 3 n ) = π 2 + m π n arg ( z 3 ) = π 2 + m π 7 30 π n = π 2 + m π n = 15 7 + 30 7 m \begin{align*}
\arg (z_3^n) &= \frac{\pi}{2} + m\pi
\\ n \arg (z_3) &= \frac{\pi}{2} + m\pi
\\ \frac{7}{30} \pi n &= \frac{\pi}{2} + m\pi
\\ n &= \frac{15}{7} + \frac{30}{7} m
\end{align*} arg ( z 3 n ) n arg ( z 3 ) 30 7 πn n = 2 π + mπ = 2 π + mπ = 2 π + mπ = 7 15 + 7 30 m
Smallest positive integer value of
n {n} n
occurs when
m = 3. {m=3.} m = 3. z 3 n = ( 3 e 7 30 π i ) 15 = ( 3 ) 14 3 e 7 2 π i = 2187 3 e − 1 2 π i \begin{align*}
z_3^n &= \left( \sqrt{3} \mathrm{e}^{\frac{7}{30} \pi \mathrm{i}} \right)^{15}
\\ &= (\sqrt{3})^{14} \sqrt{3} \mathrm{e}^{\frac{7}{2}\pi \mathrm{i}}
\\ &= 2187 \sqrt{3} \mathrm{e}^{- \frac{1}{2} \pi \mathrm{i}}
\end{align*} z 3 n = ( 3 e 30 7 π i ) 15 = ( 3 ) 14 3 e 2 7 π i = 2187 3 e − 2 1 π i ∣ z 3 n ∣ = 2187 3 ■ arg ( z 3 n ) = − 1 2 π ■ \begin{align*}
\left| z_3^n \right| &= 2187 \sqrt{3} \; \blacksquare
\\ \arg \left( z_3^n \right) &= - \frac{1}{2} \pi \; \blacksquare
\end{align*} ∣ z 3 n ∣ arg ( z 3 n ) = 2187 3 ■ = − 2 1 π ■ Question Commentary
Parts (a) and (b) are (hopefully) standard work with complex numbers in modulus-argument
form as well as the drawing of Argand diagrams.
The "purely imaginary" concept in part (c) comes hot on the heels of many schools
setting questions in a similar vein for school exams (and last seen in our national
exams in 2020 Paper 1 Question 4
and 2014 Paper 2 Question 4 ).
The use of the table in the GC can also be really helpful to determine the least
value of n {n} n