When finding the conjugate of a binomial, you just reverse the sign. So the conjugate of 3+4i is 3-4i.
The additive inverse of 6+4i is -6-4i since their sum is 0. It is analogous to real numbers where the additive inverse of 6 is -6 since 6+-6 =6-6=0 In the case of complex numbers, we add them by adding the real parts and then adding the imaginary parts. So to find the complex additive inverse of a+bi, we find the inverse of a which is -a and of bi which is -bi and so the additive inverse is -a-bi
Add the real and the imaginary parts separately.
A product is a binary operatoin. That is, it requires two numbers to be combined. There is only one number, 2 + 4i, in the question.
The concept of conjugate is usually used in complex numbers. If your complex number is a + bi, then its conjugate is a - bi.
When finding the conjugate of a binomial, you just reverse the sign. So the conjugate of 3+4i is 3-4i.
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The conjugate of -8-4i is -8+4i. It is obtained by changing the sign of the imaginary part of the complex number.
To get the conjugate simply reverse the sign of the complex part. Thus conj of 7-4i is 7+4i
4i(-2 -3i) = 4i×-2 - 4i×-3i = -8i -12i² = -8i + 12 = 12 -8i → the conjugate is 12 + 8i
The conjugate of 6 + i is 6 - i.
(2 + 4i) - (7 + 4i) = -5 2 + 4i - 7 + 4i = -5 + 8i
Since the imaginary parts cancel, and the real parts are the same, the sum is twice the real part of any of the numbers. For example, (5 + 4i) + (5 - 4i) = 5 + 5 + 4i - 4i = 10.
the problem: what is 4 + 4i + 4 + 6i what you do is add the real and imaginary parts, thus: 4+4 and 4i+6i = 8+10i answer.
To solve this type of problem, multiply both the numerator and denominator by the conjugate of the denominator. (2 - 4i) / (4 + 2i) = (2 - 4i)(4 - 2i) / (4 + 2i)(4 - 2i) then expand all the terms, and simplify. = (8 - 20i + 8i2) / (16 - 4i2) = (8 - 20i - 8) / (16 + 4) = -20i / 20 = -i Which in the required answer format becomes, 0 + i.
(x - 4i)(x + 4i) where i is the square root of -1
The additive inverse of 6+4i is -6-4i since their sum is 0. It is analogous to real numbers where the additive inverse of 6 is -6 since 6+-6 =6-6=0 In the case of complex numbers, we add them by adding the real parts and then adding the imaginary parts. So to find the complex additive inverse of a+bi, we find the inverse of a which is -a and of bi which is -bi and so the additive inverse is -a-bi