Question

find the modulus and argument of 1+i√3

Answer 1

1+iroot3 modulus means just changing all signs into positive I hope it helps you

Answer 2

Answer:

The modulus of $1+\sqrt{3i}$ is 2 and the argument of the given complex number is $60^{0}$.

Step-by-step explanation:

Given complex number is $1+\sqrt{3i}$

$z=1+\sqrt{3i}=a+ib$

Where $a=1$ and $b=\sqrt{3}$

Then the modulus of z is $\sqrt{a^2+b^2}$

$|z|=\sqrt{1^2+\sqrt{3}^2 }$

The square of the square root of 3 is 3.

$|z|=\sqrt{1+3 }$

Perform addition in the square root.

$|z|=\sqrt{4 }$

Square root of 4 is 2.

$|z|=2 }$

Therefore the modulus of given complex number is 2.

Now, Let us find the argument of given complex number is:

∅ $=tan^{-1}( \frac{b}{a} )$

Substitute the values of a and b in the above formula.

  $=tan^{-1}( \frac{\sqrt{3} }{1} )$

  $=tan^{-1}( \sqrt{3} )$

  $=60^{0}$

Therefore, the argument of the given complex number $1+\sqrt{3i}$ is $60^{0}$.