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

o aapko aara hai kya answer ?

soln for iⁱ question

i= (e^{iÎ /2})ⁱ =e^{-Î /2}

12] \large z=\frac{\sqrt{3}+i}{2}

\large \left(z^{101}+i^{103} \right)^{105}=?

a]z
b]z²
c]z³
d]z⁴

yeh hi toh answer hai arey A aur B poche hai :P

1+ω+ω2=0 therefore ge =(-ω2)7
=-ω14 =-ω2
=1+ω=A+Bω

i was able to olve till that..wt after that????

integration definite hai.........i mean she'z correct...............shall i post whole solution???????

for 1t Q

1+ω+ω²=0 therefore ge =(-ω²)⁷
=-ω¹⁴ =-ω²
=1+ω=A+Bω

A=B=1

amplitude matlab angle na tan wala

wat kya?

kya???????????????????????????????????????????????????????????

arctan(3/(4+√3)
=27.626°
wats da ans?

options hai kya

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 --------------