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The sum of a complex number and its conjugate?
Given a complex number z = a + bi, the conjugate z* = a - bi, so z + z*= a + bi + a - bi = 2*a. Note that a and b are both real numbers, and i is the imaginary unit: +sqrt(-1).
Find the exact value of cos 195?
cos(195) = -0.965925826289
What is a quadrantal angle?
A Quadrantal angle is an angle that is not in Quadrant I. Consider angle 120. You want to find cos(120) . 120 lies in quadrant II. Also, 120=180-60. So, it is enough to find cos(60) and put the proper sign. cos(60)=1/2. Cosine is negative in quadrant II, Therefore, cos(120) = -1/2.
What is this expression as the cosine of an angle cos30cos55 plus sin30sin55?
cos(30)cos(55)+sin(30)sin(55)=cos(30-55) = cos(-25)=cos(25) Note: cos(a)=cos(-a) for any angle 'a'. cos(a)cos(b)+sin(a)sin(b)=cos(a-b) for any 'a' and 'b'.
Express all the trignometric ratios in terms of COS A?
Provided that any denominator is non-zero, sin = sqrt(1 - cos^2)tan = sqrt(1 - cos^2)/cos sec = 1/cos cosec = 1/sqrt(1 - cos^2) cot = cos/sqrt(1 - cos^2)
What is the answer of double derivative of sine z to the power 6?
y = sin6(z) dy/dz = 6*sin5(z)*cos(z) then d2y/dz2 = 6*5*sin4(z)*cos(z) + 6*sin5(z)*(-sin(z)) = 6*sin4(z)*[5*cos(z) - sin2(z)]
If z equals a plus ib then show that arg conjugate of z equals 2pi -arg z?
If z = a + ib then arg(z) = arctan(b/a) Let z' denote the conjugate of z. Therefore, z' = a - ib Then arg(z') = arctan(-b/a) = 2*pi - arctan(b/a) = 2*pi - arg(z)
The sum of a complex number and its conjugate?
Given a complex number z = a + bi, the conjugate z* = a - bi, so z + z*= a + bi + a - bi = 2*a. Note that a and b are both real numbers, and i is the imaginary unit: +sqrt(-1).
What is the multiplicative inverse of 4 plus i?
To find the multiplicative inverse of a complex number z = (a + bi), divide its complex conjugate z* = (a - bi) by z* multiplied by z (and simplify): z = 4 + i z* = 4 - i multiplicative inverse of z: z* / (z*z) = (4 - i) / ((4 - i)(4 + i) = (4 - i) / (16 + 1) = (4- i) / 17 = 1/17 (4 - i)
What are the 3 complex roots of -1 using the DeMoivre's theorem?
Problem: find three solutions to z^3=-1. DeMoivre's theorem is that (cos b + i sin b)^n = cos bn + i sin bn So we can set z= (cos b + i sin b), n = 3 cos bn + i sin bn = -1. From the last equation, we know that cos bn = -1, and sin bn = 0. Three possible solutions are bn=pi, bn=3pi, bn=5pi. This gives three possible values of b: b=pi/3 b=pi b = 5pi/3. Now using z= (cos b + i sin b), we can get three possible cube roots of -1: z= (cos pi/3 + i sin pi/3), z= (cos pi + i sin pi), z= (cos 5pi/3 + i sin 5pi/3). Working these out gives -1/2+i*sqrt(3)/2 -1 -1/2-i*sqrt(3)/2
Find the conjugate of 84 - 63i?
21
Put 3-3i 3 in polar form?
3 - 3i Let's try to represent the given complex number in the polar form. z = |z|(cos θ + i sin θ) let z = 3 - 3i in the form a + bi, where a = |z|cos θ and b = |z| sin θ so that |z| = √(a2 + b2) = √[(3)2 + (- 3)2] = √(9 + 9) = √18 = 3/√2 cos θ = a/|z| = 3/3√2 = 1/√2 and sin θ = b/|z| = -3/3√2 = -1/√2 z = 3 - 3i = |z|(cos θ + i sin θ) =3√2(1/√2 - i 1/√2)
What is the complex conjugate of a plus bi?
The complex conjugate of a+bi is a-bi. This is written as z* where z is a complex number. ex. z = a+bi z* = a-bi r = 3+12i r* = 3-12i s = 5-6i s* = 5+6i t = -3+7i = 7i-3 t* = -3-7i = -(3+7i)
Find the complex conjugate of 14 plus 12i?
To find the complex conjugate of a number, change the sign in front of the imaginary part. Thus, the complex conjugate of 14 + 12i is simply 14 - 12i.
How do you find the missing parts of an oblique triangle which 3 sides are given?
Let the sides be x, y, z. Let the angles opposite those sides be X, Y, Z You can use the Cosine Law which states cos X = (y^2 + z^2 - x^2)/2yz Then calculate cos^-1(or arccos X) and this will give you the angle in degrees. then do the same for Y cos Y = (x^2 + z^2 - y^2)/2xz Do the same to get Y. Then add X and Y and subtract for 180° and you have your three angles.
Construction of two random variables such that their correlation coefficient equals -1?
y = cos ( x ). z = cos ( x + 180 degrees ).
Is the product of two conjugate complex number always a real number?
Yes. This can be verified by using a "generic" complex number, and multiplying it by its conjugate: (a + bi)(a - bi) = a2 -abi + abi + b2i2 = a2 - b2 Alternative proof: I'm going to use the * notation for complex conjugate. Any complex number w is real if and only if w=w*. Let z be a complex number. Let w = zz*. We want to prove that w*=w. This is what we get: w* = (zz*)* = z*z** (for any u and v, (uv)* = u* v*) = z*z = w
