complex numbers

{\color{red}Given \ z=\frac{\sqrt3+i}{2}}

\Longrightarrow z=cos(\frac{\pi}{6})+isin(\frac{\pi}{6}) \\ z^{102}=cos(\frac{102\pi}{6})+isin(\frac{102\pi}{6}) \\ z^{102}=-1 \Longrightarrow z^{101}=\frac{-1}{z} \\ \text{Subsitute in the given expression} \\ (\frac{-1}{z}-i)^{105} \\ \\ \text{Take minus common} \\ \Longrightarrow (-1) (\frac{1+zi}{z})^{105} \\ \text{Substitute the value of z,After simplification,we will get} \ (-1)z^{105} \\ \Longrightarrow (-1)z^{102}z^{3} \\ \Longrightarrow z^3

aacha koi ni try d way i id may b m makin sum cal mistake
z=(i+√3 )/2 zi= w^2..............
.. (or w.........m confused :( )

so then (z¹⁰¹ +i¹⁰³ )¹⁰⁵

isse solve karo

nahiin ji sry. nahiin aaara. magar answer toh z³ hi hain

nahiin ji answer z³ hain

soln for iⁱ question

i= (e^{iÎ /2})ⁱ =e^{-Î /2}

z^2

Hey Matie, inte : PL. see c this :

http://targetiit.com/iit_jee_forum/posts/sum1_pl_explain_this_2me_3321.html

hey sorry re doin
i wrte 21 on ma paper n took taht as 4 woh 4 ki atrh dikh ra tha

A= 1 AUR B=1 tapan tujhe?

sry

2] if ω is a complex cube root of unity, and

\large \left(1+\omega \right)^{7}=A+B\omega

then A and B are respectively --------------

is it

arc tan(3/(4+√3))