Rd sharma solution of chapter 13 Complex numbers

Leave a Comment / Class 11 / By Tanishk

Complex numbers

Exercise 13.1

QUESTION 1

Evaluate the following:

(i) $i^{457}$

Sol :

$=i^{4\times114+1}$

$=(i^{4})^{114} \times{i}$

$=i$ $[\because~i^4=1 \text{ and } 1^n=1]$

(ii) $i^{528}$

Sol :

$=i^{4\times132}$

$=(i^{4})^{132}$

=1 $[\because~i^4=1 \text{ and } 1^n=1]$

(iii) $\dfrac{1}{i^{158}}$

Sol :

$=\dfrac{1}{i^{4\times39+2}}$

$=\dfrac{1}{(i^{4})^{14}\times{i^2}}$

$=\dfrac{1}{i^2}$ $[\because~i^4=1\text{ and } 1^n=1]$

$=\dfrac{1}{-1}$ $[\because i^2=-1]$

= -1

(iv) $i^{37}+\dfrac{1}{i^{67}}$

Sol :

$=i^{4\times9+1}+\dfrac{1}{(i^{4})^{16+3}}$

$=(i^4)^{9}\times{i}+\dfrac{1}{(i^{4})^{16}\times{i^3}}$

$={i}+\dfrac{1}{i^3}$ $[\because~i^4=1\text{ and } 1^n=1]$

$=i+\dfrac{1}{-i}$ $[\because~i^3=-i]$

$=i-\dfrac{1}{i}\times{\dfrac{i}{i}}$

$=i-\dfrac{i}{i^2}$

$=i-(-i)$ $[\because~i^2=-1 ]$

$=2i$

(v) $({i^{41}+\dfrac{1}{i^{257}}})^9$

Sol :

$=({i^{4\times10+1}+\dfrac{1}{(i^{4})^{64+1}}})^9$

$=({(i^4)^{10}\times{i}+\dfrac{1}{(i^{4})^{64}\times{i}}})^9$

$=({{i}+\dfrac{1}{i}})^9$ $[\because~i^4=1\text{ and } 1^n=1]$

$=({i+\dfrac{1}{i}\times{\dfrac{i}{i}}})^9$

$=({i+\dfrac{i}{i^2}})^9$

$=({i+(-i)})^9$ $[\because~i^2=-1 ]$

= 0

(vi) $(i^{77}+i^{70}+i^{87}+i^{414})^3$

Sol :

$=\{i^{4\times19+1}+i^{4\times17+2}+i^{4\times21+3}+i^{4\times103+2}\}$

$=\{(i^4)^{19}\times i+(i^4)^{17}\times i^2+(i^4)^{21}\times i^3+(i^4)^{103}\times i^2\}$

$=(i+i^2+i^3+i^2)$ $[\because~i^4=1 \text{ and } 1^n=1]$

$=i-1-i-1$ $[\because~(i^2=-1)~(i^3=-i) ]$

= 0

(vii) $i^{30}+i^{40}+i^{60}$

Sol :

$=i^{4\times7+2}+i^{4\times10}+i^{4\times15}$

$=(i^4)^{7}\times i^2+(i^4)^{10}+(i^4)^{15}$

$=i^2+1+1$ $[\because~i^4=1 \text{ and } 1^n=1]$

$=-1+1+1$ $[\because~(i^2=-1)]$

= 1

(viii) $i^{49}+i^{68}+i^{89}+i^{110}$

Sol :

$=i^{4\times12+1}+i^{4\times17}+i^{4\times22+1}+i^{4\times27+2}$

$=(i^4)^{12}\times i +(i^4)^{17}+(i^4)^{22}\times i+(i^4)^{27}\times i^2$

$=i+1+i+i^{2}$ $[\because~i^4=1 \text{ and } 1^n=1]$

$=2i$ $[\because~i^{2}=-1]$

QUESTION 2

Show that $1+i^{10}+i^{20}+i^{30}$

Sol :

$=1+i^{4\times2+2}+i^{4\times5}+i^{4\times7+2}$

$=1+(i^4)^{2}\times i^2+(i^4)^{5}+(i^4)^{7}\times i^2$

$=1+i^2+1+i^2$ $[\because~i^4=1 \text{ and } 1^n=1]$

$=1-1+1-1$ $\because~i^2=-1 \text{ and }$

= 0 which is a real number

Hence, proved

QUESTION 3

Find the values for the following expressions:

(i) $i^{49}+i^{68}+i^{89}+i^{110}$

Sol :

$=i^{4\times12+1}+i^{4\times17}+i^{4\times22+1}+i^{4\times27+2}$

$=(i^4)^{12}\times i +(i^4)^{17}+(i^4)^{22}\times i+(i^4)^{27}\times i^2$

$=i+1+i+i^{2}$ $[\because~i^4=1 \text{ and } 1^n=1]$

$=2i$ $[\because~i^{2}=-1]$

(ii) $i^{30}+i^{80}+i^{120}$

Sol :

$=i^{4\times7+2}+i^{4\times20}+i^{4\times30}$

$=(i^4)^{7}\times i^2+(i^4)^{20}+(i^4)^{30}$

$=i^2+1+1$ $[\because~i^4=1 \text{ and } 1^n=1]$

$=-1+1+1$ [\because i^2=-1]

= 1

(iii) $i+i^2+i^3+i^4$

Sol :

$=i-1-i+1$ $\because~(i^2=-1)~(i^3=-i)~(i^4=1)$

= 0

(iv) $i^{5}+i^{10}+i^{15}$

Sol :

$=i^{4\times1+1}+i^{4\times2+2}i^{4\times3+3}$

$=(i^4)\times i+(i^4)^2\times i^2+(i^4)^3\times i^3$

$=i+i^2+i^3$ $[\because~i^4=1 \text{ and } 1^n=1]$

$=i-1-i$ [\because~(i^2=-1)~(i^3=-i)]

= -1

(v) $\dfrac{i^{592}+i^{590}+i^{588}+i^{586}+i^{584}}{i^{582}+i^{580}+i^{578}+i^{576}+i^{574}}$

Sol :

$=\dfrac{i^{4\times148}+i^{4\times147+2}+i^{4\times147}+i^{4\times146+2}+i^{4\times146+}}{i^{4\times145+2}+i^{4\times145}+i^{4\times144+2}+i^{4\times144}+i^{4\times143+2}}$

$=\dfrac{(i^4)^{148}+\{(i^4)^{147}\times i^2\}+\{(i^4)^{146}\}+\{(i^4)^{146}\times i^2\}+\{(i^4)^{146}\}}{\{(i^4)^{145}\times i^2\}+\{(i^4)^{145}\}+\{(i^4)^{144}\times i^2\}+\{(i^4)^{144}\}+\{(i^4)^{143}\times i^2\}}$

$=\dfrac{1+i^2+1+i^2+1}{i^2+1+i^2+1+i^2}$ $[\because~i^4=1 \text{ and } 1^n=1]$

$==\dfrac{1-1+1-1+1}{-1+1-1+1-1}$ $\because~(i^2=-1)\text{ and } (i^4=1)$

= -1

(vi) $1+i^2+i^4+i^6+i^8+....+i^{20}$

Sol :

$\because i^2=-1$

$i^4=1$

$i^6=-1$

$i^8=1$

.

.

.

$i^{20}=1$

$\therefore~1+i^2+i^4+i^6+i^8.....+i^{20}$

$=[1+(-1)]+[1+(-1)]+[1+(-1)]+.......[1+(-1)]+1$

$=5\times[1+(-1)]+1$ [As, there are 11 terms]

$=5\times0+1$

= 1

(vii) $(1+i)^6+(1-i)^3$

Sol :

$=[(1+i)^2]^3+(1-i)^3$

$=(1+i^2+2i)^3+[1^3-i^3-3\times 1\times i (1+i)]^3$

$=(2i)^3+[1+i-3i(1-i)]^3$ $(i^2=-1)\text{ and } (i^3=-i)$

$=(2i)^3+[-2-2i]$

$=8i^3-2-2i$

$=-8i-2-2i$

$=-10i-2$

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