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It depends on what you're trying to do with the complex numbers, and what level of math understanding that you are at. Some tips:

Treat the i like a variable (like x): example: Add a + bx & c + dx = a + bx + c + dx = a + c + bx + dx = (a + c) + (b + d)x. Now, substitute x = i

Multiplying: (a + bx) * (c + dx) = ac + adx + bcx + bdx2 = ac + (ad + bc)x + bdx2, when substituting x = i in this one: ac + (ad + bc)i + bdi2, but i2 = -1, so we have:

ac + (ad + bc)i - bd = (ac - bd) + (ad + bc)i

If you are familiar with vectors, you can treat complex numbers as vectors in the complex plane, and do some operations on them that way. See related link.

Q: How do you figure out complex number problems?

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A triangle, with one of the complex numbers represented by a line from the origin to the number, and then move from that point up and over the amount of the next complex number. Then draw a line segment from the origin to the final point.

Yes. And since Real numbers are a subset of complex numbers, a complex number can also be a pure real.Another AnswerYes, for example: (0 + j5) is a complex number, whose 'real' number is zero.

Graphically, the conjugate of a complex number is its reflection on the real axis.

One is a complex number and a real number.

When a complex number is multiplied by its conjugate, the product is a real number and the imaginary number disappears.

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The ratio of two quantities need not itself be a rational number - it can be a complex number.

Yes. If the number is like, for example, 3+0i, then you'll figure out that the number, though is written as a complex number, is actually a real number 'cause 0i=0 and 3+0=3 so you have both real and complex number. Every number is a complex number, no matter if it's imaginary or real or a combination of both (a+bi).

i is equalled to the square root of -1. The square root is a number which, multiplied by itself, will give a number (eg: 2 is the square root of 4 because 2x2=4) But negative numbers can't have square roots because any number multiplied by itself will be positive. So 'i' is imaginary. You use 'i' in calculus to figure out complex problems.

3 and 5 are both complex numbers, and if you multiply them together, you get 15, which is a real number. If you were looking for two non-real complex numbers, then any pair of complex conjugates will work. For example, 5+2i times 5-2i is 29.

I don't think so for 5000 years of organized governmentshave had the opportunity to fix things but have failed miserably?

A triangle, with one of the complex numbers represented by a line from the origin to the number, and then move from that point up and over the amount of the next complex number. Then draw a line segment from the origin to the final point.

Figures that can be subdivided into simple figures.

You study.

31

Adjoint operator of a complex number?

Calculators, probably in the 1980's (I know for a fact the HP 48 calculator circa 1992 handled complex and imaginary numbers) helped people perform calculations with complex numbers, without having to figure conjugates, angles, etc. on paper. Complex number computing was long before that. I know that FORTRAN (developed by IBM and released in 1957) could handle complex number calculations. FORTRAN was designed specifically for scientific and engineering calculations. Check out the Wikipedia article.

Anywhere between 0 square feet and approx 998.2 square feet - it all depends upon the exact shape of the complex figure. To calculate the area of a complex figure, split it up into shapes for which you can workout the area and then add all the areas of the shapes together.