1
AIEEE 2004
MCQ (Single Correct Answer)
+4
-1
If $$\,\left| {{z^2} - 1} \right| = {\left| z \right|^2} + 1$$, then z lies on :
A
an ellipse
B
the imaginary axis
C
a circle
D
the real axis
2
AIEEE 2003
MCQ (Single Correct Answer)
+4
-1
If $$z$$ and $$\omega $$ are two non-zero complex numbers such that $$\left| {z\omega } \right| = 1$$ and $$Arg(z) - Arg(\omega ) = {\pi \over 2},$$ then $$\,\overline {z\,} \omega $$ is equal to
A
$$- i$$
B
1
C
- 1
D
$$i$$
3
AIEEE 2003
MCQ (Single Correct Answer)
+4
-1
Let $${Z_1}$$ and $${Z_2}$$ be two roots of the equation $${Z^2} + aZ + b = 0$$, Z being complex. Further , assume that the origin, $${Z_1}$$ and $${Z_2}$$ form an equilateral triangle. Then :
A
$${a^2} = 4b$$
B
$${a^2} = b$$
C
$${a^2} = 2b$$
D
$${a^2} = 3b$$
4
AIEEE 2003
MCQ (Single Correct Answer)
+4
-1
If $${\left( {{{1 + i} \over {1 - i}}} \right)^x} = 1$$ then :
A
x = 2n + 1, where n is any positive integer
B
x = 4n , where n is any positive integer
C
x = 2n, where n is any positive integer
D
x = 4n + 1, where n is any positive integer.
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