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Q: Find the complex number that is farthest away from the complex plane?

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If you are given a plane, you can always find and number of points that are not in that plane but, given anythree points there is always at least one plane that goes through all three.

The number 1 is its own square root, cube root, etc. in the real number system. To find complex roots of unity, we use the unit circle from trigonometry, coupled with the complex plane, where the x-axis is the real axis, and the y-axis is the imaginary axis. In that coordinate system, the number 1 corresponds to the point (1, 0) and the complex number 1 + 0i. Every complex number a + bi corresponds to the point (a, b) in the complex plane. To find roots of 1, we divide the unit circle up into as many sectors as the number of roots we are trying to find. For cube roots, that's 3 of course, so we divide the unit circle up into 3 sectors of 120 degrees (or 2pi/3 radians) each. So the three cube roots we want are located at the points 120 degrees around the unit circle from (1, 0). Since points on the unit circle have coordinates (cos(theta), sin(theta)), the first one we come to will be (cos(120), sin(120)) = (-1/2, Sqrt(3)/2). This point corresponds to the complex number -1/2 + (sqrt(3)/2)*i. The next point on the circle, 120 degrees from the last one, is (cos(240), sin(240)) = (-1/2, - sqrt(3)/2) = -1/2 - (sqrt(3)/2)*i. Now you have the three cube roots of unity: 1, -1/2 + (sqrt(3)/2)*i, and -1/2 - (sqrt(3)/2)*i. There's much more to all this, involving something called DeMoivre's Formula or Theorem.

another point

yes, you can find a real root to the cube root of any negative real number. There will also be two complex roots which satisfy it, as well.

The square of a "normal" number is not negative. Consequently, within real numbers, the square root of a negative number cannot exist. However, they do exist within complex numbers (which include real numbers)and, if you do study the theory of complex numbers you wil find that all the familiar properties are true.

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This is best done if the complex number is in polar coordinates - that is, a distance from the origin, and an angle. Take the square root of the argument (the absolute value) of the complex number; and half the angle.

The answer depends on what group or field the function is defined on. In the complex plane, the range is the complex plane. If the domain is all real numbers and the radical is an odd root (cube root, fifth root etc), the range is the real numbers. Otherwise, it is the complex plane. If the domain is non-negative real numbers, the range is also the real numbers.

It is +1

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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.

you count the number of boxes IN the shape.

There is no such thing; you seem to have misunderstood something.Any real number can be regarded as a complex number with zero imaginary part, eg.: 5 = 5+0i

The number of pieces for an RC Plane Engine vary on the model of the plane. Here's a link that you may find helpful: http://www.rcairplaneswarehouse.com/

Changing a complex number from standard form (x + iy) to normal form r(Cos(theta)+isin(theta)) is relatively simple if you're comfortable with trigonometry and Pythagoras's theorem. It is easiest to imagine it in the context of the complex plane.If you were to draw your complex number on the complex plane, theta would be the angle between the number and the positive x axis, and r would be the length of the line. r is easiest to find; simply put x and y into the equation Sqrt(x2 + Y2)and the result will be r, the length of the complex number.To find theta, you must picture a triangle, imagining the length y to be the opposite, and the length x to be the adjacent, and performing arctan(y/x) to find theta - however, be careful as depending on what quadrant the number is in, you may have to perform further operations in order to find the true angle.

If you understand what the absolute value of a complex number is, skip to the tl;dr part at the bottom. The absolute value can be thought of as a sorts of 'norm', because it assigns a positive value to a number, which represents that number's "distance" from zero (except for the number zero, which has an absolute value of zero). For real numbers, the "distance" from zero is merely the number without it's sign. For complex numbers, the "distance" from zero is the length of the line drawn from 0 to the number plotted on the complex plane. In order to see why, take any complex number of the form a + b*i, where 'a' and 'b' are real numbers and 'i' is the imaginary unit. In order to plot this number on a complex plane, just simply draw a normal graph. The number is located at (a,b). In order to determine the distance from 0 (0,0) to our number (a,b) we draw a triangle using these three points: (0,0) (a,0) (a,b) Where the points (0,0) and (a,b) form the hypotenuse. The length of the hypotenuse is also the "distance" of a + b*i from zero. Because the legs run parallel to the x and y axes, the lengths of the two legs are 'a' and 'b'. By using the Pythagorean theorem, we can find the length of the hypotenuse as (a2 + b2)(1/2). Because the length of the hypotenuse is also the 'distance' of the complex number from zero on the complex plane, we have the definition: |a + b*i| = (a2 + b2)(1/2) ALRIGHT, almost there. tl;dr: Remember that the complex conjugate of a complex number a + b*i is a + (-b)*i. By plugging this into the Pythagorean theorem, we have: b2 = (-b)2 So: (a2 + (-b)2)(1/2) = (a2 + b2)(1/2) QED.

If you are given a plane, you can always find and number of points that are not in that plane but, given anythree points there is always at least one plane that goes through all three.

There are a number of websites where one can find plane games for a cell phone. They can be downloaded from the Apple Store or Google Play Store as well as 'MPBus'.

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