Jamia Millia Islamia MCA Complex Number Previous Year Questions (PYQs) – Page 3 of 3

$|Z - 1| = |Z + 1|$ represents the perpendicular bisector of the line joining $(1,0)$ and $(-1,0)$, i.e. the $y$–axis. For points on the $y$–axis, $Z = i y$. Now $|Z - i| = |i y - i| = |i(y - 1)| = |y - 1|$. Also $|Z + 1| = \sqrt{1 + y^2}$. So $\sqrt{1 + y^2} = |y - 1| \Rightarrow y = 0$. Thus only one point satisfies — $Z = 0$.

We know $\omega^3 = 1$ and $1 + \omega + \omega^2 = 0$. So $1 + \omega = -\omega^2$. Hence $(1 + \omega)^7 = (-\omega^2)^7 = (-1)^7 \omega^{14} = -\omega^{14}$. Now $\omega^{14} = \omega^{3 \times 4 + 2} = \omega^2$. Therefore $(1 + \omega)^7 = -\omega^2 = 1 + \omega$. So comparing with $A + B\omega$, we have $A = 1$, $B = 1$. $\Rightarrow A + B = 2.$

Using triangle inequality,

$$|z_1 + z_2 + 3 + 4i| \le |z_1| + |z_2| + |3 + 4i| = 4 + 3 + 5 = 12.$$

Hence, value is less than 12.

$\ \sqrt{-9} = \sqrt{9}\sqrt{-1} = 3i$

$1+\omega+\omega^2=0 \Rightarrow 1+\omega-\omega^2=2\omega.$ $(2\omega)^7 = 128\omega^7 = 128\omega.$

Conjugates ⇒ $\cos 2x = \sin x \Rightarrow x = (n + \tfrac{1}{2})\pi.$

Diagonals bisect each other ⇒ $z_1 + z_3 = z_2 + z_4.$