Rd sharma solution of chapter 13 Complex numbers

TanishkClass 11

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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