Download App

Maths Case study (4 Marks)

Complex Numbers and Quadratic Equations · Uttarakhand Board (UBSE) STD 11 Science (Uttarakhand - English Medium). 3 questions.

Questions

Case study (4 Marks)

🎯

Test yourself on this topic

3 questions · timed · auto-graded

Question 14 Marks
We have, $i=\sqrt{-1}$. So, we can write the higher powers of $i$ as follows
(i) $i^2=-1$
(ii) $i^3=i^2 \cdot i=(-1) \cdot i=-i$
(iii) $i^4=\left(i^2\right)^2=(-1)^2=1$
(iv) $i^5=i^{4+1}=i^4 \cdot i=1 \cdot i=i$
(v) $i^6=i^{4+2}=i^4 \cdot i^2=1 \cdot i^2=-1$

In order to compute $i^n$ for $n>4$, write $i^n=i^{4 q+r}$ for some $q, r \in N$ and $0 \leq r \leq 3$. Then, $i^n=$ $i^{4 q} \cdot i^r=\left(i^4\right)^q \cdot i^r=(1)^q \cdot i^r=i^r$.
In general, for any integer $k, i^{4 k}=1, i^{4 k+1}=i, i^{4 k+2}=-1$ and $i^{4 k+3}=-i$.

On the basis of above information, answer the following questions.

(i) The value of $i^{37}$ is equal to
    (a) $i$     (b) $-i$     (c) 1     (d) -1

(ii) The value of $i^{-30}$ is equal to
    (a) $i$     (b) 1     (c) -1     (d) $-i$

(iii) If $z=i^9+i^{19}$, then $z$ is equal to
    (a) $0+0 i$     (b) $1+0 i$     (c) $0+i$     (d) $1+2 i$

(iv) The value of $\left[i^{19}+\left(\frac{1}{i}\right)^{25}\right]^2$ is equal to
    (a) -4     (b) 4     (c) $\mathrm{i}$     (d) 1

(v) If $z=i^{-39}$, then simplest form of $z$ is equal to
    (a) $1+0 i$     (b) $0+i$     (c) $0+0 i$     (d) $1+i$
Answer
$\text { (i) - (a); (ii) - (c); (iii) - (a); (iv) - (a); (v) - (b) }$
View full question & answer→
Question 24 Marks
A complex number $z$ is pure real if and only if $\bar{z}=z$ and is pure imaginary if and only if $\bar{z}=-z$.
Based on the above information, answer the following questions.

(i) If $(1+i) z=(1-i) \bar{z}$, then $-i \bar{z}$ is
(a) $-\bar{z}$     (b) $z$     (c) $\bar{z}$     (d) $z^{-1}$

(ii) $\overline{Z_1 Z_2}$ is
(a) $\bar{z}_1 \bar{z}_2$     (b) $\bar{z}_1+\bar{z}_2$     (c) $\frac{z_1}{z_2}$     (d) $\frac{1}{z_1 z_2}$

(iii) If $x$ and $y$ are real numbers and the complex number $\frac{(2+i) x-i}{4+i}+\frac{(1-i) y+2 i}{4 i}$ is pure real, the relation between $x$ and $y$ is
(a) $8 x-17 y=16$          (b) $8 x+17 y=16$
(c) $17 x-8 y=16$          (d) $17 x-8 y=-16$

(iv) If $z=\frac{3+2 i \sin \theta}{1-2 i \sin \theta}\left(0<\theta \leq \frac{\pi}{2}\right)$ is pure imaginary, then $\theta$ is equal to
(a) $\frac{\pi}{4}$     (b) $\frac{4}{6}$     (c) $\frac{6}{3}$     (d) $\frac{\pi}{12}$
(v) If $z_1$ and $z_2$ are complex numbers such that $\left|\frac{z_1-z_2}{z_1+z_2}\right|=1$
(a) $\frac{z_1}{z_2}$ is pure real          (b) $\frac{z_1}{z_2}$ is pure imaginary
(c) $z_1$ is pure real          (d) $z_1$ and $z_2$ are pure imaginary
Answer
$(i)-(b) ;(i)-(a) ;(i i)-(a) ;(i v)-(c) ;(v)-(b)$
View full question & answer→
Question 34 Marks
Two complex numbers $Z_1=a+i b$ and $Z_2=c+i d$ are said to be equal, if $a=c$ and $b=d$.
On the basis of above information, answer the following questions.

(i) If $(3 a-6)+2 i b=-6 b+(6+a) i$, then the real values of $a$ and $b$ are respectively
    (a) $-2,2$     (b) $2,-2$     (c) $3,-3$     (d) 4,2

(ii) If $(2 a+2 b)+i(b-a)=-4 i$, then the real values of $a$ and $b$ are respectively.
    (a) 2,3     (b) $2,-2$     (c) 3,1     (d) $-2,2$

(iii) If $\left(\frac{1-i}{1+i}\right)^{100}=a+i b$, then the values of $a$ and $b$ are respectively
    (a) 1,0     (b) 0,1     (c) 1,2     (d) 2,1

(iv) If $\frac{(1+i)^2}{2-i}=x+i y$, then the value of $x+y$ is
    (a) $\frac{1}{5}$     (b) $\frac{3}{5}$     (c) $\frac{4}{5}$     (d) $\frac{2}{5}$

(v) If $(x+y)+i(x-y)=4+6 i$, then $x y$ is equal to
    (a) 5     (b) -5     (c) 4     (d) -4
Answer
$\text { (i) - (a); (ii)-(b); (iii)-(a); (iv)-(d); (v)-(b) }$
View full question & answer→
Generate Paper FreeAll Complex Numbers and Quadratic Equations topics
Case study (4 Marks) - Maths STD 11 Science Questions - Vidyadip