Question

Find modulus and argument of a complex number 1-i root 3

Answer 1

Answer:

-2Π/3

Step-by-step explanation:

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Answer 2

Answer:

Modulus of 1-i√3 = 2

Argument of 1-i√3  = $\frac{5\pi }{3}$

Step-by-step explanation:

To find,

The modulus and argument of the complex number 1-i√3

Recall the formula

The modulus of the complex number x+iy = $\sqrt{x^2+y^2}$

The argument of the complex number x+iy  = $tan^{-1}({\frac{y}{x})$

Solution:

The complex number given is 1-i√3

Comparing with x+iy we get x = 1 and y = -√3

Modulus of 1-i√3

The modulus of the complex number 1-i√3= $\sqrt{1^2+(-\sqrt{3})^2 }$

= $\sqrt{1+3 }$

=$\sqrt{4}$

= 2

∴ Modulus of 1-i√3 = 2

Argument of 1-i√3

Since Argument of  x+iy  = $tan^{-1}({\frac{y}{x})$ we have

Arugment of 1-i√3 = $tan^{-1}({\frac{-\sqrt{3} }{1})$

We need to find the angle 'θ' such that tan θ = -$\sqrt{3}$

The value of tanθ is negative, in the second and fourth quadrants

Since here the value of y is negative and x is positive, the angle 'θ' is in the fourth quadrant

θ = 2π - $tan^{-1}(\sqrt{3})$ = 2π - $\frac{\pi }{3}$ = $\frac{5\pi }{3}$

Argument of 1-i√3 = $tan^{-1}({\frac{-\sqrt{3} }{1})$ = $\frac{5\pi }{3}$

∴Modulus of 1-i√3 = 2

Argument of 1-i√3  = $\frac{5\pi }{3}$

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