JEE Advanced 2017 Paper 1 Offline | Complex Numbers Question 16 | Mathematics

JEE Advanced 2017 Paper 1 Offline

MCQ (More than One Correct Answer)

-1

Let a, b, x and y be real numbers such that a $$-$$ b = 1 and y $$ \ne $$ 0. If the complex number z = x + iy satisfies $${\mathop{\rm Im}\nolimits} \left( {{{az + b} \over {z + 1}}} \right) = y$$, then which of the following is(are) possible value(s) of x?

$$1 - \sqrt {1 + {y^2}} $$

$$ - 1 - \sqrt {1 - {y^2}} $$

$$1 + \sqrt {1 + {y^2}} $$

$$ - 1 + \sqrt {1 - {y^2}} $$

JEE Advanced 2016 Paper 2 Offline

MCQ (More than One Correct Answer)

-2

Let $$a,\,b \in R\,and\,{a^{2\,}} + {b^2} \ne 0$$. Suppose
$$S = \left\{ {Z \in C:Z = {1 \over {a + ibt}}, + \in R,t \ne 0} \right\}$$, where $$i = \sqrt { - 1} $$. Ifz = x + iy and z $$ \in $$ S, then (x, y) lies on

the circle with radius $${{1 \over {2a}}}$$and centre $$\left\{ {{1 \over {2a}},\,0} \right\}\,for\,a > 0\,,b \ne \,0$$

the circle with radius $$-{{1 \over {2a}}}$$and centre $$\left\{ -{{1 \over {2a}},\,0} \right\}\,for\,a < 0\,,b \ne \,0$$

the x-axis for $$a \ne \,\,0,\,b \ne \,0$$

the y-axis for $$a = \,\,0,\,b \ne \,0$$

JEE Advanced 2013 Paper 2 Offline

MCQ (More than One Correct Answer)

-1

Let $\omega=\frac{\sqrt{3}+i}{2}$ and $P=\left\{\omega^n: n=1,2,3, \ldots\right\}$. Further

$\mathrm{H}_1=\left\{z \in \mathrm{C}: \operatorname{Re} z<\frac{1}{2}\right\}$ and

$\mathrm{H}_2=\left\{z \in \mathrm{C}: \operatorname{Re} z<\frac{-1}{2}\right\}$, where C is the

set of all complex numbers. If $z_1 \in \mathrm{P} \cap \mathrm{H}_1, z_2 \in$ $\mathrm{P} \cap \mathrm{H}_2$ and O

represents the origin, then $\angle z_1 \mathrm{O} z_2=$

$${\pi \over 2}$$

$${\pi \over 6}\,$$

$${{2\pi } \over 3}$$

$${{5\pi } \over 6}$$

IIT-JEE 2010 Paper 1 Offline

MCQ (More than One Correct Answer)

-1

Let $${{z_1}}$$ and $${{z_2}}$$ be two distinct complex number and let z =( 1 - t)$${{z_1}}$$ + t$${{z_2}}$$ for some real number t with 0 < t < 1. IfArg (w) denote the principal argument of a non-zero complex number w, then

$$\left| {z - {z_1}} \right| + \left| {z - {z_2}} \right| = \left| {{z_1} - {z_2}} \right|$$

Arg $$(z - {z_1})$$ = Arg$$(z - {z_2})$$

$$\left| {\matrix{ {z - {z_1}} & {\overline z - {{\overline z }_1}} \cr {{z_2} - {z_1}} & {{{\overline z }_2} - {{\overline z }_1}} \cr } } \right|$$ = 0

Arg $$(z - {z_1})$$ = Arg$$({z_2} - {z_1})$$

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