Question

find the modulus and argument of the complex number minus root 3 + I

Answer 1

3+1=4 and it is square root of 2

Answer 2

Answer:

ordered pair (x, y) = -√3,

-√3=r cos thita and 1=r sin thita

by squaring and adding both sides, we get

(-√3) ^2+1^2=r^2cos^2thita r^2sin^2 thita

3+1=r^2(cos^2thita +sin^2thita)

4=r^2(1)

r=√4=2

hence, modulus=r=|r|=2

therefore, cos thita =x/r= -√3/2 and sin thita =y/r=1/2

since cos thita is negative and sin thita is positive

so argument will be in third quadrant

arg(z) =(180-thita) =(180-30) =150° = 150×π/180°=5π/6

hence r=2 and arg(z) = 5π/6