If z1 = √3 - i and z2 = 1 + i √3, then amp (z1 + z2) =

Question

If z1 = √3 - i and z2 = 1 + i √3, then amp (z1 + z2) =  (A) (π/12) (B) (π/15) (C) (π/6) (D) (π/4)

Tardigrade Admin · Accepted Answer

We have , z1 = √3 - i and z2 = 1 + i √3 So, z1 +z2 =(√3 +1)+ i(√3 -1) Putting r cos θ =√3 +1 and r sin θ =√3 -1 tan θ =(√3 - 1/3+1) ×(√3 -1/√3 - 1) = ((√3 -1)2/3-1) = (3+1-2√3/2) tan θ =(4-2√3/2)= 2-√3 tan θ = tan (π/12) amplitude (θ) = (π/12)

Q. If $z_{1} = 3 - i$ and $z_{2} = 1 + i 3$ , then amp $(z_{1} + z_{2}) =$

$\frac{π}{12}$

$\frac{π}{15}$

$\frac{π}{6}$

$\frac{π}{4}$

Q. If z1​=3​−i and z2​=1+i3​, then amp (z1​+z2​)=

12π​

15π​

6π​

4π​

We have , z1​=3​−i and z2​=1+i3​ So, z1​+z2​=(3​+1)+i(3​−1) Putting rcosθ=3​+1 and rsinθ=3​−1 tanθ=3+13​−1​×3​−13​−1​ =3−1(3​−1)2​=23+1−23​​ tanθ=24−23​​=2−3​ tanθ=tan12π​ amplitude (θ)=12π​

Q. If $z_{1} = 3 - i$ and $z_{2} = 1 + i 3$ , then amp $(z_{1} + z_{2}) =$

$\frac{π}{12}$

$\frac{π}{15}$

$\frac{π}{6}$

$\frac{π}{4}$