Find the modulus and principal argument of the complex numbers giving the argument in radians either as pi or a decimal correct to 2 decimal places
c) i square root 3 -2 ALL UPON -2-i square root 3
(-2 + √3i)/(-2 - √3i)
recall that (a+bi)(a-bi) = a^2+b^2
Use that to get rid of the complex number in the bottom:
(-2+√3i)/(-2-√3i) * (-2+√3i)/(-2+√3i)
(-2+√3i)^2 / (4+3)
(4-4√3i-3)/7
1/7 - 4√3/7 i
THANKS!! now I can find the modulus and argument
=D
To find the modulus and principal argument of a complex number, we'll start by writing the complex number in the form a+bi, where a and b are real numbers.
Given complex number: (i√3 - 2) / (-2 - i√3)
To simplify the expression, let's rationalize the denominator by multiplying both the numerator and the denominator by the conjugate of the denominator.
Conjugate of -2 - i√3 = -2 + i√3
So, multiplying both the numerator and the denominator by the conjugate, we get:
[(i√3 - 2) / (-2 - i√3)] * [(-2 + i√3) / (-2 + i√3)]
Simplifying this expression, we have:
[(i√3 - 2) * (-2 + i√3)] / [(-2)^2 - (i√3)^2]
= [-2i√3 - 4 + 3i^2] / [4 + 3]
= [-2i√3 - 4 - 3] / 7
= (-2i√3 - 7) / 7
Now, let's write the result in the form a + bi:
-2i√3 / 7 - 7 / 7
= -2i√3 / 7 - 1
So, the complex number can be written as: -1 - (2i√3 / 7)
Now, let's find the modulus (absolute value) of the complex number:
|z| = √(a^2 + b^2)
Here, a = -1, b = -2√3 / 7
|z| = √((-1)^2 + (-2√3 / 7)^2)
= √(1 + 12 / 49)
= √(61 / 49)
= √61 / 7
The modulus of the complex number is √61 / 7.
Next, let's find the principal argument (in radians) of the complex number using arctan2:
θ = arctan2(b, a)
Here, a = -1, b = -2√3 / 7
θ = arctan2(-2√3 / 7, -1)
≈ -1.011 radians (rounded to 2 decimal places)
Therefore, the modulus of the complex number is √61 / 7 and the principal argument is approximately -1.011 radians.