Qus : 1
🎓 AMU MCA 📅 Year: 2019 📚 Mathematics 🏷 Complex Number
3
A real value of $x$ will satisfy the equation $\left(\dfrac{3 - 4ix}{3 + 4ix}\right)=\alpha - i\beta; (\alpha,\beta \text{ are real}),$ if
1
$\alpha^2 + \beta^2 = -1$
2
$\alpha^2 - \beta^2 = 1$
3
$\alpha^2 + \beta^2 = 1$
4
$\alpha^2 - \beta^2 = 1$
✓ Solution
Qus : 2
🎓 AMU MCA 📅 Year: 2019 📚 Mathematics 🏷 Complex Number
3
If $\omega$ is a complex cube root of unity, then the value of
$\dfrac{p + q\omega + r\omega^2}{r + p\omega + q\omega^2} + \dfrac{p + q\omega + r\omega^2}{q + r\omega + p\omega^2}$ $(p,q,r \in R)$ is equal to
✓ Solution
Qus : 3
🎓 AMU MCA 📅 Year: 2019 📚 Mathematics 🏷 Complex Number
2
If $P, Q, R, S$ are represented by the complex numbers $4+i, 1+6i, -4+3i, -1-2i$ respectively, then $PQRS$ is a
✓ Solution
Qus : 4
🎓 AMU MCA 📅 Year: 2019 📚 Mathematics 🏷 Complex Number
3
If $n$ is a positive integer, then $(1+i)^n + (1-i)^n$ is equal to
1
$(\sqrt{2})^{n-2}\cos\left(\dfrac{n\pi}{4}\right)$
2
$(\sqrt{2})^{n-2}\sin\left(\dfrac{n\pi}{4}\right)$
3
$(\sqrt{2})^{n+2}\cos\left(\dfrac{n\pi}{4}\right)$
4
$(\sqrt{2})^{n+2}\sin\left(\dfrac{n\pi}{4}\right)$
✓ Solution
Qus : 5
🎓 AMU MCA 📅 Year: 2018 📚 Mathematics 🏷 Complex Number
2
The amplitude of the complex number $1+\sin\alpha-i\cos\alpha$ is
3
$\frac{\alpha}{2}-\frac{\pi}{4}$
✓ Solution
Qus : 6
🎓 AMU MCA 📅 Year: 2018 📚 Mathematics 🏷 Complex Number
2
The points $z_1=x+iy$ and $z_2=\frac{1}{-x+iy}$ in the complex plane lie on
1
a circle with centre origin
2
a straight line through origin
✓ Solution
Qus : 7
🎓 AMU MCA 📅 Year: 2025 📚 Mathematics 🏷 Complex Number
1
The ring $Z[\sqrt{3}] = {a + b\sqrt{3} ; a,b \in Z}$
3
has units in the form $a + 0\sqrt{3}$ i.e., $Z$
✓ Solution
$Z[\sqrt{3}]$ has no zero divisors and is closed under addition and multiplication, hence it is an integral domain.
Qus : 8
🎓 AMU MCA 📅 Year: 2016 📚 Mathematics 🏷 Complex Number
4
The value of $\sum_{n=1}^{13}(i^n + i^{n+1})$, where $i=\sqrt{-1}$ equals
✓ Solution
$i^n$ repeats every 4 terms
Pair terms cancel ⇒ sum reduces to $i-1$
Final Answer: $\boxed{i-1}$
Qus : 9
🎓 AMU MCA 📅 Year: 2016 📚 Mathematics 🏷 Complex Number
1
The value of $\left(\frac{1+i}{1-i}\right)^{100}$ is equal to
✓ Solution
$\frac{1+i}{1-i} = i$
So expression = $i^{100} = (i^4)^{25} = 1$