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All of them. Real numbers are a subset of complex numbers.

Q: Which of these are complex numbers 5 3i 1 2i?

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When adding and subtracting complex numbers, you can treat the "i" as any variable. For example, 5i + 3i = 8i, 5i -3i = 2i, etc.; (2 + 5i) - (3 - 3i) = (2 - 3) + (5 + 3)i = -1 + 8i.

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0+3i has a complex conjugate of 0-3i thus when you multiply them together (0+3i)(0-3i)= 0-9i2 i2= -1 0--9 = 0+9 =9 conjugates are used to eliminate the imaginary parts

5+6i , -2-2i , 100+i.A complex number consists of a real part and an imaginary part: a+bi where 'i' is the imaginary unit (sq.rt(-1)).

A "complex number" is a number of the form a+bi, where a and b are both real numbers and i is the principal square root of -1. Since b can be equal to 0, you see that the real numbers are a subset of the complex numbers. Similarly, since a can be zero, the imaginary numbers are a subset of the complex numbers. So let's take two complex numbers: a+bi and c+di (where a, b, c, and d are real). We add them together and we get: (a+c) + (b+d)i The sum of two real numbers is always real, so a+c is a real number and b+d is a real number, so the sum of two complex numbers is a complex number. What you may really be wondering is whether the sum of two non-real complex numbers can ever be a real number. The answer is yes: (3+2i) + (5-2i) = 8. In fact, the complex numbers form an algebraic field. The sum, difference, product, and quotient of any two complex numbers (except division by 0) is a complex number (keeping in mind the special case that both real and imaginary numbers are a subset of the complex numbers).

Related questions

if you mean both dimensions are complex numbers, then you use foil. Example (1+i)(1+2i)= 1 + 3i - 2 (since i2 = -1) -1+3i that's a rectangle but you should understand if your in a class with complex #

(-2 + 3i) + (-1 - 2i) = -2 + 3i - 1 - 2i = -2 - 1 + 3i - 2i = -3 + i

When adding and subtracting complex numbers, you can treat the "i" as any variable. For example, 5i + 3i = 8i, 5i -3i = 2i, etc.; (2 + 5i) - (3 - 3i) = (2 - 3) + (5 + 3)i = -1 + 8i.

To multiply complex numbers you can use the same FOIL rule that you use for multiplying binomials (First, Inside, Outside, Last).(4 - 3i)(5 + 2i) = (4)(5) +(4)(2i) - (3i)(5) - (3i)(2i) = 20 + 8i-15i - 6(i)^2= 20 -7i - 6(-1) = 20 + 6 -7i = 26 -7i.

this is a very good question. lets solve (2+3i)/(4-2i). we want to make 4-2i real by multiplying it by the conjugate, or 4+2i (4-2i)(4+2i)=16-8i+8i+4=20, now we have (2+3i)/20 0r 1/10 + 3i/20 notice that -2i times 2i = -4i^2 =-4 times -1 = 4

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(1+i)3 = 1 + 3i - 3 - i = -2 + 2i This is a complex number, and therefore cannot be plotted on a Cartesian plane.

The four roots are:1 + 2i, 1 - 2i, 3i and -3i.

The question has no answer in real numbers. The solution, in complex numbers, are 2+3i and 2-3i where i is the imaginary square root of -1.

Imaginary number is a number that consist of only Imaginary part. Such as i, 40i, 1/2i, etc. While the difference between the imaginary numbers and the complex numbers are that complex number also contains Real numbers, and can be written as a + bi. For example, 30+i, 1/2+1/2i, etc.

x2 + 9 = (x + 3i)(x - 3i) The answer is based on a knowledge of imaginary and complex numbers where i2 = -1

When dividing complex numbers you must:Write the problem in fractional formRationalize the denominator by multiplying the numerator and denominator by the conjugate of the denominator.You must remember that a complex number times its conjugate will give a real number.a complex number 2+2i. the conjugate to this is 2-i1. Multiply both together gives a real number.(2+2i)(2-2i) = 4 -4i + 4i + (-4i2) (and as i2 = -1) = 8To divide a complex number by a real number simply divide the real parts by the divisor.(8+4i)/2 = (4+2i)To divide a real number by a complex number.1. make a fraction of the expression 8/(2+2i)2. multiply by 1. express 1 as a fraction of the divisor's conjunction. 8/(2+2i)*(2-2i)/(2-2i)3. multiply numerator by numerator and denominator by denominator.(16-16i)/84. and simplify 2-2i