Question

# The unit vector making an angle $$60^o$$ with the $$x-$$axis can be expressed as complex number by

A
eiπ3
B
e2iπ3
C
e4iπ3
D
all the above

Solution

## The correct option is D all the aboveGiven:-A unit vector making an angle $${ 60 }^{ o }$$with $$x−axis.$$Solution:- We have to represent the unit vector in complex . Complex form$$Z={ r }{ e }^{ i\theta }$$Which is the modulus $$\& \theta$$ is argument.So we can write as.$$\left\{ r=1 \right\}$$$$Z={ e }^{ i\theta }\left\{ \theta ={ \pi }/{ 3 } \right\} ={ 60 }^{ o }$$$$Z={ e }^{ i \dfrac { \pi }{ 3 } }$$Let $$\theta =\cfrac { \pi }{ 3 } ,\cfrac { 2\pi }{ 3 } ,\cfrac { 4\pi }{ 3 }$$ case$$(1)=\theta =\cfrac { \pi }{ 3 } =a\left\{ Where\cfrac { \pi }{ 3 } lies\quad in\quad { 1 }^{ st }\quad equation \right\}$$ so argument$$\theta =a$$ Hence$$=Z={ e }^{ i\theta }={ e }^{ i\dfrac { \pi }{ 3 } }$$ Case$$(2)=a=\cfrac { 2\pi }{ 3 } \left\{ Where\cfrac { 2\pi }{ 3 } lies\quad in\quad { 2 }^{ nd }\quad equation \right\}$$So argument$$\theta =\pi -a$$$$=\pi -\cfrac { 2\pi }{ 3 }$$$$=\cfrac { \pi }{ 3 }$$  Hence$$z={ e }^{ i\dfrac { \pi }{ 3 } }$$ Case$$(3)=a=\cfrac { 4\pi }{ 3 } \left\{ Where\cfrac { 4\pi }{ 3 } lies\quad in\quad { 3 }^{ rd }\quad equation \right\}$$ So argument$$\theta =\pi +a$$$$=\pi +\cfrac { 4\pi }{ 3 }$$$$=\cfrac { 7\pi }{ 3 } =2\pi +\cfrac { \pi }{ 3 }$$ So argument $$\tan\theta =\tan\left( 2\pi +\cfrac { \pi }{ 3 } \right) =\tan\cfrac { \pi }{ 3 }$$ Hence$$z={ e }^{ i\dfrac { \pi }{ 3 } }$$ Hence,unit vector can be represented as complex no.$$={ e }^{ i\dfrac { \pi }{ 3 } },{ e }^{ i\dfrac { 2\pi }{ 3 } },{ e }^{ i\dfrac { 4\pi }{ 3 } }$$Maths

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