What is the trigonometric notation for 3-3(sqrt3i)?
Hmmm. The i is in the sqrt( )? that means the sqrt of the sqrt if -1.
Is this in the complex plane?
I am not certain what you are doing.
the law of cosine. I am stuck on the issue of converting 3-3 sqare root 3i, and am not sure how to do this.
for example square root 3-i = 2(cos 33 + i sin 330)
To find the trigonometric notation for the complex number 3-3(sqrt(3)i), we can use the polar form of a complex number.
The polar form of a complex number is given by the expression r(cosθ + isinθ), where r is the modulus (magnitude) of the complex number and θ is the argument (angle) in radians.
First, let's find the modulus of the complex number. The modulus can be found using the formula |z| = sqrt(a^2 + b^2), where a and b are the real and imaginary parts of the complex number, respectively.
In this case, a = 3 and b = -3(sqrt(3)). Therefore, the modulus is given by:
|r| = sqrt(3^2 + (-3(sqrt(3)))^2)
|r| = sqrt(9 + 27)
|r| = sqrt(36)
|r| = 6
Next, let's find the argument of the complex number using the formula tanθ = b/a. Rearranging the formula, we have: θ = atan(b/a), where atan denotes the inverse tangent function.
In this case, a = 3 and b = -3(sqrt(3)). Therefore, the argument (in radians) is given by:
θ = atan(-3(sqrt(3)) / 3)
θ = atan(-sqrt(3))
θ ≈ -π/3
Now that we have the modulus and argument, we can express the complex number in trigonometric notation.
3-3(sqrt(3)i) can be written as:
6(cos(-π/3) + isin(-π/3))
So, the trigonometric notation for 3-3(sqrt(3)i) is 6(cos(-π/3) + isin(-π/3)).