Register

ISI2020-MMA: 2

• edited by
225 views
0 votes
0 votes

Let $z = (1 – t^{2}) + i \sqrt{1 - t^{2}}$ be a complex number where $t$ is a real number such that $|t| < 1$. Then the locus of $z$ in the complex plane is

  1. An ellipse

  2. A hyperbola

  3. A parabola

  4. A pair of straight lines

• edited by
User Avatar

Please log in or register to answer this question.

Voting & Accepting an answer helps improve the community
← PreviousNext →
← Previous in categoryNext in category →

Related questions

0 votes
0 votes
1 answer
1
admin asked Jul 23, 2022
338 views
ISI2020-MMA: 1
Let $\text{A} = (1, -1), \text{B} = (-2, 0), \text{C} = (1, 2)$ and $\text{D}$ be the vertices of a parallelogram in the $\text{X – Y}$ plane listed clockwise. Then the point $\text{D}$ is $(4, 1)$ $(-2, -3)$ $(3, 0)$ $(-2, 1)$
Let $\text{A} = (1, -1), \text{B} = (-2, 0), \text{C} = (1, 2)$ and $\text{D}$ be the vertices of a parallelogram in the $\text{X – Y}$ plane listed clockwise. Then the...
User Avatar
0 votes
0 votes
0 answers
2
admin asked Jul 23, 2022
288 views
ISI2020-MMA: 3
Let $\int ^{2}_{1} e^{x^{2}} dx = a$. Then the value of $\int ^{e^{4}}_{e} \sqrt{\log_{e} x} dx$ is $e^{4} - a$ $2e^{4} - a$ $e^{4} - e – 4a$ $2e^{4} - e – a$
Let $\int ^{2}_{1} e^{x^{2}} dx = a$. Then the value of $\int ^{e^{4}}_{e} \sqrt{\log_{e} x} dx$ is$e^{4} - a$$2e^{4} - a$$e^{4} - e – 4a$$2e^{4} - e – a$
User Avatar
0 votes
0 votes
0 answers
3
admin asked Jul 23, 2022
246 views
ISI2020-MMA: 4
The area bounded by the curves $y = e^{x}, y = xe^{x}$ and the $y$ - axis is $e – 2$ $e + 2$ $e – 1$ $2e – 3$
The area bounded by the curves $y = e^{x}, y = xe^{x}$ and the $y$ - axis is$e – 2$$e + 2$$e – 1$$2e – 3$
User Avatar
0 votes
0 votes
1 answer
4
admin asked Jul 23, 2022
316 views
ISI2020-MMA: 5
The set of all solutions of the inequality $\frac{1}{2^{x} - 1} > \frac{1}{1 - 2^{x - 1}}$ is. $\left(1, \infty \right)$ $\left(0, \log_{2} \left ( \frac{4}{3} \right )\right)$ $\left(0, \log_{2} \left ( \frac{4}{3} \right )\right) \cup \left(1, \infty \right)$ $\left(-1, \infty \right)$
The set of all solutions of the inequality $\frac{1}{2^{x} - 1} \frac{1}{1 - 2^{x - 1}}$is.$\left(1, \infty \right)$$\left(0, \log...
User Avatar
Link Copied!