Mathematics Class XI

Question 1 of 25

1. Question
1 point(s)
If $ w $ is a cube root of unity and $(1+w)^7=A+Bw $, the $ A $ and $ B $ are respectively_____
- $0, 1 $
- $ 1, 1$
- $ 1, 0$
- $-1, 1 $
Correct

Incorrect

Hint

$\begin{aligned} (1+w)^7&=(1+w)(1+w)^6\\&=(1+w)(-w^2)^6=1+w \end{aligned}$
$\implies A+Bw=1+w$
$\implies A=1, ~B=1$
Question 2 of 25

2. Question
1 point(s)
The value of $ \sum_{k=1}^{6}\left( \sin \dfrac{2\pi k}{7}-i \cos \dfrac{2 \pi k}{7} \right)$ is______
- $-1$
- $ 0 $
- $ -i$
- $ i$
Correct

Incorrect

Hint

$\sum_{k=1}^{6}\left( \sin \dfrac{2 \pi k}{7}-i\cos \dfrac{2 \pi k}{7} \right)$
$=\sum_{k=1}^{6}-i\left( \cos \dfrac{2k\pi}{7}+\sin \dfrac{2k \pi}{7} \right)$
$=-i\left\{ \sum_{k=1}^{6}~e^{\dfrac{i2k\pi}{7}} \right\}$
$=-i\left\{ e^{\dfrac{i2\pi}{7}}+ e^{\dfrac{i4\pi}{7}}+ e^{\dfrac{i6\pi}{7}}+ e^{\dfrac{i8\pi}{7}}+ e^{\dfrac{i10\pi}{7}}+ e^{\dfrac{i12\pi}{7}}\right\}$
$=-i \left\{ e^{\dfrac{i2\pi}{7}}\dfrac{\left( 1-e^{\dfrac{i2\pi}{7}} \right)}{1-e^{\dfrac{i2\pi}{7}}} \right\}$
$=-i\left\{ \dfrac{e^{\dfrac{i2\pi}{7}}-e^{\dfrac{i4\pi}{7}}}{1-e^{\dfrac{i2\pi}{7}}} \right\}$
$=-i\left( \dfrac{e^{\dfrac{i2\pi}{7}}-1}{1-e^{\dfrac{i2\pi}{7}}} \right)=i$
Question 3 of 25

3. Question
1 point(s)
Let $Z_k=\cos \left( \dfrac{3 k \pi}{10} \right)+i \sin \left( \dfrac{2k \pi}{10} \right); k=1,2,…9 $. For each $Z_k $, there exists a $Z_j $ such that $ Z_k \cdot Z_j$ is equal to______
- $0 $
- $-1 $
- $ 1$
- $2 $
Correct

Incorrect

Hint

Given, $Z_k=e^{i\dfrac{2k \pi}{10}}$
$\implies Z_k \cdot Z_j=e^{i\left( \dfrac{2\pi}{10} \right)(k+j)}$
$Z_k$ is $100^{\text {th}}$ root of unity.
Taking, $Z_j$ as $\overline {Z_k}$ , we have $Z_k\cdot \overline {Z_k}=1$
$\implies Z_k\cdot \overline {Z_j}=1$
Question 4 of 25

4. Question
1 point(s)
Let $ w $ be the complex number $ \cos \dfrac{2\pi}{3}+i \sin \dfrac{2 \pi}{3}$. Then the number of distinct complex number $ z $ satisfying $ \begin{vmatrix}
z+1 & w &w^2 \\
w & z+w &1 \\
w^2 & 1 & z+w
\end{vmatrix}=0$ is equal to_____
- $0 $
- $ 3$
- $ 2$
- $1 $
Correct

Incorrect

Hint

Let $A=\begin{vmatrix} z+1 &w & w^2 \\ w & z+w &1 \\ w^2 & 1 & z+w \end{vmatrix}$
Now, $A^2=\begin{bmatrix} 0 & 0 & 0 \\ 0& 0& 0 \\ 0& 0 & 0 \end{bmatrix}, ~ |A|=0$
$\implies A^3=0$
$\implies \begin{vmatrix} z+1 &w &w^2 \\ w& z+w^2 & 1 \\ w^2 &1 & z+w \end{vmatrix}=\left[ A+Z1 \right]=0$
$\implies z^3=0$
$\implies z=0$ , the number of $z$ satisfying the given equation is $1$ .
Question 5 of 25

5. Question
1 point(s)
The value of the expression $ 1(2-w)(2-w^2)+2(3-w)(3-w^2)+…+(n-1)(n-w)(n-w^2)$ where, $ w $ is an imaginary cube root of unity is______
- $ n $
- $\dfrac{n(n+1)}{2} $
- $\left[ \dfrac{n(n+1)}{2} \right]^2-n$
- $ \left[ \dfrac{n(n-1)}{2} \right]^2$
Correct

Incorrect

Hint

$\begin{aligned}T_r&=(r-1)(r-w)(r-w^2)~~(\text {Say}) \\&=r^3-1 \end{aligned}$
$\begin{aligned} \therefore S_n&=\sum_{r=1}^{n}(r^3-1) \\&=\left[ \dfrac{n(n+1)}{2} \right]^2-n \end{aligned}$
Question 6 of 25

6. Question
1 point(s)
The cube roots of unity when represented on the Argand plane form the vertices of an or a _______
- right-angled triangle
- equilateral triangle
- Isosceles triangle
- Scalene triangle
Correct

Incorrect

Hint

Since, cube roots of unity are $1,w,w^2$ given by $A(1, 0), ~ B \left( -\dfrac{1}{2}, \dfrac{\sqrt{3}}{2} \right), ~ C\left( -\dfrac{1}{2}, -\dfrac{\sqrt{3}}{2} \right)$
$\implies AB=BC=CA=\sqrt{3}$
$\therefore ABC$ is an equilateral triangle.
Question 7 of 25

7. Question
1 point(s)
If $ x=a+b, \quad y=a \alpha+b \beta,\quad z=a\beta+b \alpha$, where $\alpha, \beta $ are complex cube roots of unity, then $ xyz$ is equal to_____
- $a^2+b^2 $
- $ a^3+b^3$
- $a^2-b^2 $
- $a^3-b^3 $
Correct

Incorrect

Hint

Let $\alpha=w \text { and } \beta=w^2$
$\begin{aligned}xyz&=(a+b)(a\alpha+b \beta)(a\beta+b \alpha)\\&=(a+b)\left[ a^2 \alpha \beta+ab(\alpha^2+\beta^2)+b^2 \alpha \beta \right]\\&=(a+b)\left[ a^2(w\cdot w^2)+ab(w^2+w^4)+b^2(w \cdot w^2) \right]\\&=(a+b)(a^2-ab+b^2)\left[ \therefore 1+w+w^2=0 \text { and } w^3=1 \right]\\&=a^3+b^3 \end{aligned}$
Question 8 of 25

8. Question
1 point(s)
For $n \in N $, find the value of $ (1-i)^n\left( 1-\dfrac{1}{i} \right)^n$.
- $ 2^{n+1} $
- $ 2^n$
- $2^{n-1} $
- None of these
Correct

Incorrect

Hint

$\begin{aligned} (1-i)^n\left( 1-\dfrac{1}{i} \right)^n&=(1-i)^n(i-1)^n\cdot (i)^{-n}\\&=(1-i)^n(1-i)^n(-1)^n(i)^{-n}\\&=\left[ (1-i)^2 \right]^n(-1)^n(i)^{-n}\\&=(1+i^2-2i)^n(-1)^n(i)^{-n} \\&=(1-1-2i)^n(-1)^n(i)^{-n}\\&=(-2)^ni^n(-1)^n(i)^{-n}\\&=(-1)^{2n}\cdot 2^n=2^n\end{aligned}$
Question 9 of 25

9. Question
1 point(s)
Solve: $\sum_{n=1}^{13}\left( i^n+i^{n+1} \right) $, where $ n \in N $.
- $ i+1$
- $i-1 $
- $i $
- $ -i$
Correct

Incorrect

Hint

$\begin{aligned}\sum_{n=1}^{13}\left( i^n+i^{n+1} \right) &=\sum_{n=1}^{13} ~i^n(1+i)\\&=(1+i)\left( i+i^2+i^3+...+i^{13} \right)\\&=(1+i)(i^{13})\left[ \therefore i^n+i^{n+1}+i^{n+2}+i^{n+3}=0 \text{ where } n \in N \text { i.e.,}\sum_{n=1}^{12}~ i^n=0 \right] \\&=(1+i)\left( i^{12}\cdot i \right)\\&=(1+i)(i^4)^3 \cdot i\\&=(1+i)i\\&=(i+i^2)\\&=i-1 \end{aligned}$
Question 10 of 25

10. Question
1 point(s)
If $ \dfrac{(1+i)^2}{2-i}=x+iy$, then find the value of $ x+y $.
- $ \dfrac{4}{5}$
- $\dfrac{1}{5} $
- $\dfrac{2}{5} $
- $\dfrac{3}{5} $
Correct

Incorrect

Hint

$\dfrac{(1+i)^2}{2-i}=x+iy$
$\implies \dfrac{1+i^2+2i}{2-i}=x+iy$
$\implies \dfrac{2i}{2-i}=x+iy$
$\implies \dfrac{2i(2+i)}{(2-i)(2+i)}=x+iy \implies \dfrac{4i+2i^2}{4-i^2}=x+iy$
$\implies \dfrac{4i-2}{4+1}=x+iy\implies -\dfrac{2}{5}+\dfrac{4i}{5}=x+iy$
$\implies x=-\dfrac{2}{5}, \quad y=\dfrac{4}{5}$
$x+y=-\dfrac{2}{5}+\dfrac{4}{5}=\dfrac{2}{5}$
Question 11 of 25

11. Question
1 point(s)
If $\left( \dfrac{1-i}{1+i} \right)^{100}=a+ib $, then find $ (a, b)$.
- $ (0, 1)$
- $(1, 1) $
- $(1, 0)$
- $(0,-1) $
Correct

Incorrect

Hint

$\left( \dfrac{1-i}{1+i} \right)^{100}=a+ib$
$\implies \left[ \dfrac{(1-i)(1-i)}{(1+i)(1-i)} \right]^{100}=a+ib$
$\implies \left( \dfrac{1+i^2-2i}{1-i^2} \right)^{100}=a+ib$
$\implies \left( -\dfrac{2i}{2} \right)^{100}=a+ib$
$\implies (-1)^{100}(i^4)^{25}=a+ib$
$\implies 1=a+ib$
Hence $a=1, ~b=0$
Question 12 of 25

12. Question
1 point(s)
If $a=\cos \theta+i \sin \theta $, find the value of $ \dfrac{1+a}{1-a}$.
- $-i \cot \dfrac{\theta}{2} $
- $ i \tan \dfrac{\theta}{2}$
- $ -i \tan \dfrac{\theta}{2}$
- $ i \cot \dfrac{\theta}{2}$
Correct

Incorrect

Hint

$a=\cos \theta+i \sin \theta$
$\dfrac{1+a}{1-a}=\dfrac{1+\cos \theta+i \sin \theta}{1-\cos \theta-i \sin \theta}$
$=\dfrac{1+2\cos^2 \dfrac{\theta}{2}-1+2i \sin \dfrac{\theta}{2}\cdot \cos \dfrac{\theta}{2}}{1-1+2\sin^2 \dfrac{\theta}{2}-2i \sin \dfrac{\theta}{2}\cdot \cos \dfrac{\theta}{2}}$
$=\dfrac{2 \cos \dfrac{\theta}{2}\left( \cos \dfrac{\theta}{2}+ i \sin \dfrac{\theta}{2} \right)}{2 \sin \dfrac{\theta}{2}\left( \sin \dfrac{\theta}{2}+i \sin \dfrac{\theta}{2} \right)}$
$=\dfrac{2 \cos \dfrac{\theta}{2}\left( \cos \dfrac{\theta}{2}+i \sin \dfrac{\theta}{2} \right)}{2 i \sin \dfrac{\theta}{2}\left( \cos \dfrac{\theta}{2}+i \sin\dfrac{\theta}{2} \right)}$
$=-\dfrac{1}{i}\cot \dfrac{\theta}{2}=\dfrac{i^2}{i}\cot \dfrac{\theta}{2}$
$=i \cot \dfrac{\theta}{2}$
Question 13 of 25

13. Question
1 point(s)
If $ (1+i)z=(1-i)\overline {z}$, then $ z $ equals to_______
- $\overline{z} $
- $i\overline{z} $
- $ -i\overline{z}$
- $ -(\overline{z})$
Correct

Incorrect

Hint

$(1+i)z=(1-i)\overline{z}\implies \dfrac{z}{\overline{z}}=\dfrac{1-i}{1+i}$
$\implies \dfrac{z}{\overline{z}}=\dfrac{(1-i)(1-i)}{(1+i)(1-i)}=\dfrac{1+i^2-2i}{1-i^2}$
$\implies \dfrac{z}{\overline{z}}=\dfrac{1-1-2i}{2}=\dfrac{-2i}{2}=-i$
$\implies z=-i \overline{z}$
Question 14 of 25

14. Question
1 point(s)
If $z=x+iy $, where $ z \overline{z}+2(z+\overline{z})+b=0$, $b \in R $, represents_____
- a circle
- an ellipse
- a parabola
- a hyperbola
Correct

Incorrect

Hint

$z=x+iy \implies \overline{z}=x-iy$
$z\overline{z}+2(z+\overline{z})+b=0$
$\implies (x+iy)(x-iy)+2(x+iy+x-iy)+b=0$
$\implies x^2+y^2+4x+b=0$ , which is the equation of a circle.
Question 15 of 25

15. Question
1 point(s)
If $ \dfrac{z-1}{z+1}$ is a purely imaginary number, then find the value of $ |z| (z \neq -1)$
- $ 0 $
- $ 1$
- $2 $
- None of these
Correct

Incorrect

Hint

$z=x+iy$
$\dfrac{z-1}{z+1}=\dfrac{x+iy-1}{x+iy+1}=\dfrac{(x-1+iy)(x+1-iy)}{(x+1+iy)(x+1-iy)}$
$=\dfrac{(x+1)(x-1)+iy(x+1)-iy(x-1)-i^2y^2}{(x+1)^2-(iy)^2}$
$\implies \dfrac{z-1}{z+1}=\dfrac{x^2-1+y^2+1\left[ y(x+1)-y(x-1) \right]}{(x+1)^2+y^2}$
Since $\dfrac{z-1}{z+1}$ is purely imaginary number then $Re\left( \dfrac{z-1}{z+1} \right)=0$
$\implies \dfrac{(x^2-1)+y^2}{(x+1)^2+y^2}=0$
$\implies x^2-1+y^2=-0 \implies x^2+y^2=1$
$\implies \sqrt{x^2+y^2}=\sqrt{1}\implies |z|=1$
Question 16 of 25

16. Question
1 point(s)
If $z_1 \text { and } z_2 $ are two complex numbers such that $ |z_1|=|z_2|$ and $\arg (z_1) +\arg (z_2)=\pi $, then _____
- $ z_1=\overline{z_2}$
- $z_1=-\overline{z_2} $
- $ z_1=\dfrac{\overline{z_2}}{2}$
- $z_1=-\dfrac{\overline{z_2}}{2} $
Correct

Incorrect

Hint

Let $z_1=r_1(\cos \theta_1+i \sin \theta_1) \text { and } z_2=r_2(\cos \theta_2+i \sin \theta_2)$
$|z_1|=|z_2|\text { and } \arg (z_1)+\arg (z_2)=\pi$
If $|z_1|=|z_2| \implies r_1=r_2$ if $\arg (z_1)+\arg (z_2)=\pi$
$\implies \theta_1+\theta_2=\pi \implies \theta_1=\pi-\theta_2$
Now, $z_1=r_1 (\cos \theta_1+i \sin \theta_2)$
$\begin{aligned}\implies z_1&=r_2\left[ \cos (\pi-\theta_2)+i \sin (\pi-\theta_2) \right]\\&=r_2(-\cos \theta_2+i \sin \theta_2)\\&=-r_2(\cos \theta_2-i \sin \theta_2)\\&=-\left[ r_2(\cos \theta_2-i \sin \theta_2) \right] \end{aligned}$
$\implies z_1=-\overline {z_2}$
Question 17 of 25

17. Question
1 point(s)
If $ |z_1|=1 \text { and } z_2=\dfrac{z_1-1}{z_1+1} (z_1 \neq -1)$ , then $ Re (z_2)$ is_____
- $ 1$
- $ 2$
- $ 0$
- None of these
Correct

Incorrect

Hint

$z_1=x+iy \implies |z_1|=\sqrt{x^2+y^2=1}$
$\begin{aligned}z_2&=\dfrac{z_1-1}{z_2+1}=\dfrac{x+iy-1}{x+iy+1}=\dfrac{x-1+iy}{x+1+iy} \\&=\dfrac{(x-1+iy)(x+1-iy)}{(x+1+iy)(x+1-iy)}=\dfrac{x^2-1+ixy+iy-ixy+iy+y^2}{(x+1)^2+y^2}\\&=\dfrac{x^2+y^2-1+2iy}{(x+1)^2+y^2}=\dfrac{1-1+2iy}{(x+1)^2+y^2} \quad\left[ \therefore x^2+y^2=1 \right]\\&=0+\dfrac{2yi}{(x+1)^2+y^2} \end{aligned}$
$Re(z_2)=0$
Question 18 of 25

18. Question
1 point(s)
If $ z_1, z_2 \text { and } z_3,z_4$ are two pairs of conjugate number, then find $\arg \left( \dfrac{z_1}{z_4} \right)+\arg \left( \dfrac{z_2}{z_3} \right) $
- $ 1$
- $ 0$
- $\dfrac{1}{2} $
- None of these
Correct

Incorrect

Hint

Let $z_1=r_1(\cos \theta_1+i \sin \theta_1)$
$\begin{aligned}\text {then ,} z_2&=\overline{z_1}=r_1(\cos \theta_1-i \sin \theta_1)\\&=r_1\left[ \cos (-\theta_1)+i \sin (-\theta_1) \right] \end{aligned}$
Let $z_3=r_2(\cos \theta_2+i \sin \theta_2)$
$\text {then ,}\begin{aligned} z_4&=\overline{z_3}=r^2(\cos \theta_2-i \sin \theta_2) \\&=r_2\left[ \cos (-\theta_2)+i \sin (-\theta_2) \right] \end{aligned}$
$\arg \left( \dfrac{z_1}{z_4} \right)+\arg \left( \dfrac{z_2}{z_3} \right)=\arg (z_1)-\arg (z_4)+\arg (z_2)-\arg (z_3)$
$=\theta_1-(-\theta_2)+(-\theta_1)-\theta_2$
$=\theta_1+\theta_2-\theta_1-\theta_2=0$
Question 19 of 25

19. Question
1 point(s)
The value of $\sqrt{-25}\times \sqrt{-9} $ is______
- $15$
- $ -15$
- $15i $
- $ -15i$
Correct

Incorrect

Hint

$\sqrt{-25}\times \sqrt{-9}=\sqrt{25}~i \times \sqrt{9} ~i=5i \times3i =-15$
Question 20 of 25

20. Question
1 point(s)
Multiplicative inverse of $ 1+i $ is______
- $ 1-i$
- $\dfrac{1}{2} (1+i)$
- $\dfrac{1}{2}(1-i) $
- $ 2+i $
Correct

Incorrect

Hint

Multiplicative inverse of $1+i=\dfrac{1}{1+i}$
$=\dfrac{1-i}{(1+i)(1-i)}=\dfrac{1-i}{1-i^2}=\dfrac{1}{2}(1-i)$
Question 21 of 25

21. Question
1 point(s)
$ \arg (z)+\arg (\overline{z})$, where $ (\overline{z} \neq 0) $ is_____
- $ \theta$
- $ 0$
- $ -\theta$
- None of these
Correct

Incorrect

Hint

$\begin{aligned} \arg (z)+\arg (\overline{z})&=\theta+(-\theta) \quad \left[ \text { Let } \arg (z)=\theta \right] \\&=0 \end{aligned}$
Question 22 of 25

22. Question
1 point(s)
Solve:$\dfrac{(1-i)^3}{1-i^3} $
- $1 $
- $2 $
- $-1 $
- $-2$
Correct

Incorrect

Hint

$\begin{aligned} \dfrac{(1-i)^3}{1-i^3}&=\dfrac{(1-i)^3}{(1-i)(1+i+i^2)} =\dfrac{(1-i)^2}{(1+i+i^2)} \\&=\dfrac{1+i^2-2i}{i}=\dfrac{-2i}{i}\\&=-2\end{aligned}$
Question 23 of 25

23. Question
1 point(s)
The sum of the series $ i+i^2+i^3+…$ up to $1000$ terms____
- $ i$
- $ -i$
- $ 0$
- $1 $
Correct

Incorrect

Hint

$i+i^2+i^3+...$ up to $1000$ terms $=i+i^2+i^3+...+i^{1000}=0$
$\left[ \therefore i^n+i^{n+1}+ i^{n+2}+i^{n+3}=0, \text { where } n \in N, \sum_{i=1}^{1000}~i^n=0\right]$
Question 24 of 25

24. Question
1 point(s)
For any complex $ z $, the minimum value of $ |z|+|z-1| $ is_______
- $ 1$
- $ 2$
- $ 0$
- $-1 $
Correct

Incorrect

Hint

Let $z=x+iy$
$|z|+|z-1|=\sqrt{x^2+y^2}+\sqrt{(x-1)^2+y^2}$
If $x=0, ~ y=0$ , then value of $|z|+|z-1|=0+\sqrt{(-1)^2}=1$
Question 25 of 25

25. Question
1 point(s)
If $ z $ is a complex number such that $ z \neq 0$ and $ Re (z)=0$, then_____
- $ Im (z^2)=1$
- $ Im (z^2)=0$
- $ Im (z^2)=2y$
- $Im (z^2)=-y^2 $
Correct

Incorrect

Hint

Let $z=x+iy,~ z\neq 0,~ Re (z)=0$ i.e., $x=0, ~~z=iy$
$\therefore Im (z^2)=i^2y^2=-y^2$ , which is real.