Complex numbers: VJC 2024 Promo Q7

(a)(i)$v+2w=5\Rightarrow3v+6w=15$ ———- (1)

$3v-w^*=1+5i$ —— (2)

(1) $-$ (2): $6w+w^*=14-5i$

Let $w=x+iy$

$6(x+iy)+(x-iy)=14-5i$

$7x+5yi=14-5i$

Comparing real and imaginary parts:

$7x=14\Rightarrow x=2$

$5y=-5\Rightarrow y=-1$

Hence, $w=2-i$

Sub $w=2-i$ into $v+2w=5:\ v=5-2(2-i)=1+2i$

Alternatively

$v+2w=5\Rightarrow v=5-2w$ and sub into eqn (2):

$3(5-2w)-w^*=1+5i$

$15-6w-w^*=1+5i$

Let $w=x+yi:$

$15-6(x+yi)-(x-yi)=1+5i$

$7x+5yi=14-5i$ and compare real and imaginary parts

7(a)(ii)

7(a)(ii)Transformation: $W$ is rotated anti-clockwise through $\dfrac\pi2$ about $O$ to obtain $V$ OR $W$ is rotated clockwise through $\dfrac{3\pi}{2}$ about $O$ to obtain $V$

7(b)Since $z_1^2+z_2^2+z_3^2+z_4^2<0$, the equation has at least one non-real root. In addition, since the coefficients of the polynomial equation are all real, by conjugate root theorem, there is at least 1 pair of complex conjugate roots. Hence, at most two of $z_1$, $z_2$, $z_3$ and $z_4$ are real.