The complex number z=(i-1/ cos (π)3+i sin (π)3 is equal to :

Question

The complex number z=(i-1/ cos (π)3+i sin (π)3 is equal to : (A) √2 i( cos (5 π/12)-i sin (5 π/12)) (B)  cos (π/12)-i sin (π/12) (C) √2( cos (π/12)+i sin (π/12)) (D) √2( cos (5 π/12)+i sin (5 π/12))

Tardigrade Admin · Accepted Answer

Z =( i -1/ cos (π)3+ i sin (π)3=( i -1/(1)2+(√3)2 i =( i -1/(1)2+(√3)2 i × ((1/2)-√(3/2) i /(1)2-√3 / 2 i )=(√3-1/2)+(√3+1/2) i Apply polar form, r cos θ=(√3-1/2) r sin θ=(√3+1/2) Now, tan θ=(√3+1/√3-1) So, θ=(5 π/12)

Q. The complex number $z = \frac{i - 1}{c o s \frac{π}{3} + i s i n \frac{π}{3}}$ is equal to :

$2 i (cos \frac{5 π}{12} - i sin \frac{5 π}{12})$

$cos \frac{π}{12} - i sin \frac{π}{12}$

$2 (cos \frac{π}{12} + i sin \frac{π}{12})$

$2 (cos \frac{5 π}{12} + i sin \frac{5 π}{12})$

Q. The complex number z=cos3π​+isin3π​i−1​ is equal to :

2​i(cos125π​−isin125π​)

cos12π​−isin12π​

2​(cos12π​+isin12π​)

2​(cos125π​+isin125π​)

Z=cos3π​+isin3π​i−1​=21​+23​​ii−1​ =21​+23​​ii−1​×21​−3/2​i21​−23​i​​=23​−1​+23​+1​i Apply polar form, rcosθ=23​−1​ rsinθ=23​+1​ Now, tanθ=3​−13​+1​ So, θ=125π​

Q. The complex number $z = \frac{i - 1}{c o s \frac{π}{3} + i s i n \frac{π}{3}}$ is equal to :

$2 i (cos \frac{5 π}{12} - i sin \frac{5 π}{12})$

$cos \frac{π}{12} - i sin \frac{π}{12}$

$2 (cos \frac{π}{12} + i sin \frac{π}{12})$

$2 (cos \frac{5 π}{12} + i sin \frac{5 π}{12})$