Question

4.(a) Prove that
(−1 + i√3)^n + (−1 − i√3)^n
has either the value 2^n+1 or the value −2^n. if n is any integer (positive, negative or zero).

(b) The complex numbers z1 and z2 are connected by the relation
z1 = z2 +1/z2

If the point representing z2 in the Argand diagram describes a circle of radius a and centre at the origin, show that the point representing z1 describes the ellipse
x^2/(1 + a^2)^2 + y^2/(1 − a^2)^2=1/a^2.

Answer 1

ANSWER

(a)

Let $z=(-1+i\sqrt { 3\ })$ , then $|z|=\sqrt { 1+3 } =2,\quad \varphi =arg(z)=\frac { 2\pi }{ 3 } ,\quad \overline { z } =-1-i\sqrt { 3 }$

Angle $φ$ is in quadrant II. If $n$ is any integer, then by the formula of De Moivre, we get ${ (-1+i\sqrt { 3 } ) }^{ n }+{ (-1-i\sqrt { 3 } ) }^{ n }={ 2 }^{ n }{ \left( \cos { \frac { 2n\pi }{ 3 } \ } +i\sin { \frac { 2n\pi }{ 3 } +\cos { \frac { 2n\pi }{ 3 } } -i\sin { \frac { 2n\pi }{ 3 } \quad } } \right) = }$ $={ 2 }^{ n+1 }\cos { \frac { 2n\pi }{ 3 } } =\begin{cases} { -2 }^{ n },\quad n=3k-2 \\ -{ 2 }^{ n },\quad n=3k-1\quad \\ { 2 }^{ n+1 },\quad n=3k \end{cases}$ , because $\cos { \frac { 2(3k-2)\pi }{ 3 } } =\cos { \left( 2k\pi -\frac { 4\pi }{ 3 } \right) = } \cos { \left( \pi +\frac { \pi }{ 3 } \right) } =-\cos { \frac { \pi }{ 3 } =-\frac { 1 }{ 2 } }$ ,$\cos { \frac { 2(3k-1)\pi }{ 3 } } =\cos { \left( 2k\pi -\frac { 2\pi }{ 3 } \right) = } \cos { \left( \pi -\frac { \pi }{ 3 } \right) } =-\cos { \frac { \pi }{ 3 } =-\frac { 1 }{ 2 } }$ . So, ${ (-1+i\sqrt { 3 } ) }^{ n }+{ (-1-i\sqrt { 3 } ) }^{ n }$ has either the value ${ 2 }^{ n+1 }$ or the value $-2^n$ . if n is any

integer.

(b)If the point representing $z_2$ in the Argand diagram describes a circle of radius a and centre at the origin, then ${ z }_{ 2 }\cdot \overline { { z }_{ 2 } } =\ { |{ z }_{ 2 }| }^{ 2 }={ a }^{ 2 }$ , ${ z }_{ 2 }=a(\cos { t+i\sin { t } } ),\ \overline { { z }_{ 2 } } \ =a(\cos { t } \ -i\sin { t } )\ ,\quad { z }_{ 2 }++\frac { \overline { { z }_{ 2 } } }{ { a }^{ 2\quad } } =a(\cos { t+i\sin { t } } )++\frac { \quad (\cos { t } \ -i\sin { t } ) }{ a }$ Let ${ z }_{ 1 }=x+iy={ z }_{ 2 }+\frac { \overline { { z }_{ 2 } } }{ { a }^{ 2\ } }$ , then $x(t)=a\cos { t } +\frac { \cos { t }}{ a } ,\quad y(t)=a\sin { t-\frac { \sin { t } }{ a } }$ .

${ x }^{ 2 }(t)={ \left( a+\frac { 1 }{ a } \right) }^{ 2 }\cos ^{ 2 }{ t } =\frac { { \left( { a }^{ 2 }+1 \right) }^{ 2 }\cos ^{ 2 }{ t } }{ { a }^{ 2 } } ,\quad { y }^{ 2 }(t)={ \left( a-\frac { 1 }{ a } \right) }^{ 2 }\sin ^{ 2 }{ t= } \frac { { \left( { a }^{ 2 }-1 \right) }^{ 2 }\sin ^{ 2 }{ t } }{ { a }^{ 2 } }$ . So, $\frac { { x }^{ 2 }(t) }{ { \left( { a }^{ 2 }+1 \right) }^{ 2 } } =\frac { \cos ^{ 2 }{ t } }{ { a }^{ 2 } } ,\quad \frac { { y }^{ 2 }(t) }{ { \left( { a }^{ 2 }-1 \right) }^{ 2 } } =\frac { \sin ^{ 2 }{ t } }{ { a }^{ 2 } }$ . Adding the last two equalities , we obtain , that the point representing $z_1$ describes the ellipse $\frac { { x }^{ 2 }\quad }{ { \left( { a }^{ 2 }+1 \right) }^{ 2 } } +\frac { { y }^{ 2 }\quad }{ { \left( { 1-a }^{ 2 }\quad \right) }^{ 2 } } =\frac { 1 }{ { a }^{ 2 } }$