Question

which complex number corresponds to 2e^ipi/6

Answer 1

@sidsiddhartha

Answer 2

use
\[e^{i \theta}=\cos \theta+i \sin \theta\]

Answer 3

cospi= -1 and sin pi= 0

Answer 4

-1/3

Answer 5

nope why ure taking pi , it will be pi/6 \[ \\ \huge2e^{i \frac{ \pi }{ 6 }}=2*[\cos (\pi/6)+i~\sin (\pi/6)]\]

Answer 6

now simplify it

Answer 7

got this?

Answer 8

yes ,
2[root 3/2]+isin[1/2]

Answer 9

sorry 1/2i

Answer 10

yup so it will be
\[Z=\sqrt{3}+i\]

Answer 11

Thank you!!

Answer 12

i have another question on vector lenghts and angles

Answer 13

np :)

Answer 14

go on

Answer 15

i will have to attach a file, will do it in a min. Hang on kindly.

Answer 16

question 4 pl

Answer 17

but there is no number 4

Answer 18

sorry

Answer 19

i mean the three come under question 4

Answer 20

i need the solutions for those

Answer 21

consider any complex number-\[z=a+i~b ~~\\then ~~its~~vector~~lenght ~~is,R=\sqrt{a^2+b^2}\\and~~\angle ~~is,\theta=\tan^{-1} \frac{ b }{ a }\]

Answer 22

just use this property

Answer 23

so 3+ 4i, length is root(25)= 5

Answer 24

angle=tan inverse (4/3)

Answer 25

yup thats ok :)

Answer 26

do we need to calculate arc tan ?

Answer 27

yes u can calculate it

Answer 28

for second part is it 6+0i?

Answer 29

yes now here vector lenght,and angle is given , so
again use eulers formula which is
\[\large Z=r*e^{i \theta}=6*e^{i \frac{ \pi }{ 4 }}=6*[\cos(\pi/4)+i~\sin(\pi/4)]\]

Answer 30

Thanks Sid

Answer 31

no problem :)

Answer 32

tomorrow by this time are u free ?

Answer 33

i have doubts on Advance calculus 1

Answer 34

sorry ireally cant say but i'll free from 7 pm to (whole night ) lol

Answer 35

Thanks. let me try to hold u tomorrow. Thanks for todays session though.

Answer 36

okey ^_^