Jamia Millia Islamia Complex Number Previous Year Questions (PYQs) – Page 1 of 3

Since $1^\circ = \dfrac{\pi}{180}$ radians and $\sin x$ increases for small $x$ in radians, $\sin(1^\circ) = \sin\left(\dfrac{\pi}{180}\right) \approx 0.01745 < \sin(1 \text{ rad}) \approx 0.841$. Hence, $\sin 1^\circ < \sin 1$.

Let $z = x + iy$ and $|z| = 1$. The given condition means $(z - 1)$ is perpendicular to $(z + i)$. \[ (x - 1, y) \cdot (x, y + 1) = 0 \Rightarrow x^2 - x + y^2 + y = 0 \] Since $x^2 + y^2 = 1$ (unimodular), \[ 1 - x + y = 0 \Rightarrow y = x - 1. \] Substitute in $x^2 + y^2 = 1$: \[ x^2 + (x - 1)^2 = 1 \Rightarrow 2x^2 - 2x = 0 \Rightarrow x(x - 1) = 0. \] Hence $x = 0$ or $x = 1$. For $x = 0$, $z = -i$ (denominator $z+i=0$). For $x = 1$, $z = 1$ (numerator $z-1=0$). Both make $\arg$ undefined. Therefore, no unimodular complex number satisfies the condition.

From $10!$ onward, every factorial ends with at least two zeros. So, only the sum of first nine factorials affects the tens digit. $1! + 2! + 3! + 4! + 5! + 6! + 7! + 8! + 9! = 1 + 2 + 6 + 24 + 120 + 720 + 5040 + 40320 + 362880 = 409113.$ Tens digit = $\boxed{1}.$

Use $\omega^3 = 1$ and $1 + \omega + \omega^2 = 0$. Evaluating the determinant gives value $= -3$.

Solution: Work modulo $64$. $49\equiv -15\pmod{64}$ and $49^{1}\equiv49,\ 49^{2}\equiv33,\ 49^{3}\equiv17,\ 49^{4}\equiv1$; hence $49^{n}$ is periodic with period $4$. Also $16n\equiv 0,16,32,48\ (\bmod\ 64)$ for $n\equiv 0,1,2,3$. For each residue class: $n\equiv0$: $1+0+\lambda\equiv0\Rightarrow \lambda\equiv -1$ $n\equiv1$: $49+16+\lambda\equiv1+\lambda\equiv0$ $n\equiv2$: $33+32+\lambda\equiv1+\lambda\equiv0$ $n\equiv3$: $17+48+\lambda\equiv1+\lambda\equiv0$ All give $\lambda\equiv -1\pmod{64}$. The least negative representative is $\boxed{-1}$.

$\arg z = \tan^{-1}\!\left(\frac{1}{\sqrt3}\right)=\frac{\pi}{6}$ (I quadrant), $\arg\bar z = -\frac{\pi}{6}$. Sum of absolute values $=\frac{\pi}{6}+\frac{\pi}{6}=\frac{\pi}{3}$.

$\sqrt{3}+i = 2(\cos\tfrac{\pi}{6} + i\sin\tfrac{\pi}{6}) \Rightarrow (\sqrt{3}+i)^{17} = 2^{17}\left[\cos\tfrac{17\pi}{6} + i\sin\tfrac{17\pi}{6}\right] = 2^{16}(-\sqrt{3}+i).$ $1 - i = \sqrt{2}(\cos(-\tfrac{\pi}{4}) + i\sin(-\tfrac{\pi}{4})) \Rightarrow (1-i)^{50} = 2^{25}\left[\cos(-\tfrac{25\pi}{2}) + i\sin(-\tfrac{25\pi}{2})\right] = -i\,2^{25}.$ $\text{Ratio} = 2^{-9}\dfrac{-\sqrt{3}+i}{-i} = 2^{-9}(-1-\sqrt{3}i).$

$\dfrac{1+i}{1-i}=i$. So $i^{x}=1\Rightarrow x\equiv0\pmod4$. Only $68$ is a multiple of $4$.

$\omega^{3}=1$ and $\omega+\omega^{2}=-1$. So $1-\omega-\omega^{2}=2$ and $1+\omega^{3}=2$; product $=4$.