Not necessarily. It can be wholly imaginary.For example, 1 + i actually has two complex conjugates. Most schools will teach you that the complex conjugate is 1 - i. However, -1 + i is also a conjugate for 1 + i. (Their product is -1 times the product of the "normal" conjugate pair).The sum of 1 + i and -1 + i = 2i
The complex conjugate pair 2.5 - i*0.5*sqrt(71) and 2.5 + i*0.5*sqrt(71) where i is the imaginary number representing the square root of -1.
The two numbers are the complex conjugate pair27.5 - 79.0174iand27.5 + 79.0174iwhere i is the imaginary square root of -1.
The two numbers are the complex conjugate pair, 3.5 - 7.6i and 3.5 + 7.6i where i is the imaginary root of -1
The absolute value of the sum of two complex numbers is less than or equal to the sum of their absolute values.
Not necessarily. It can be wholly imaginary.For example, 1 + i actually has two complex conjugates. Most schools will teach you that the complex conjugate is 1 - i. However, -1 + i is also a conjugate for 1 + i. (Their product is -1 times the product of the "normal" conjugate pair).The sum of 1 + i and -1 + i = 2i
"Conjugate" usually means that in one of two parts, the sign is changed - as in a complex conjugate. If the second part is missing, the conjugate is the same as the original number - in this case, 100.
0.15 - 2.44489i and its complex conjugate, 0.15 + 2.44489i where i is the imaginary square root of -1.
The complex conjugate pair 2.5 - i*0.5*sqrt(71) and 2.5 + i*0.5*sqrt(71) where i is the imaginary number representing the square root of -1.
The two numbers are the complex conjugate pair27.5 - 79.0174iand27.5 + 79.0174iwhere i is the imaginary square root of -1.
The two numbers are the complex conjugate pair, 3.5 - 7.6i and 3.5 + 7.6i where i is the imaginary root of -1
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The absolute value of the sum of two complex numbers is less than or equal to the sum of their absolute values.
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
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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).
This might be a complex number and its conjugate: (a + bi) times (a - bi). More generally, any two complex numbers such that the angle formed by one is the negative of the angle formed by the other. In other words, you can multiply the conjugate by any real constant and still get a real result: (a + bi) times (ca - cbi). Specific examples: Multiply (3 + 2i) times (3 - 2i). Multiply (3 + 2i) times (6 - 4i).