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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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What is the conjugate of -6 plus 4i?
-6-4i.
What is the product for 2 plus 4i?
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.
What two numbers multiply to 20 but add together to 4?
24
Express the square root-48 plus 35 plus square root 25 plus square root-27 in simplest a plus bi form?
√-48 + 35 + √ 25 + √-27 = √[(i2)(16)(3)] + 35 + 5 + √[(i2)(9)(3)] = 4i√3 + 38 + 9i√3 = 38 + (13√3)i
What is the sum of (7 plus 3i) plus (8 plus 9i)?
(7 + 3i) + (8 + 9i) = (7 + 8) + (3i + 9i) = (7 + 8) + (3 + 9)i = 15 + 12i Which can also be written as: 15 + 12i = 3(5 + 4i).
What is the conjugate of -6 plus 4i?
-6-4i.
What is the conjugate of -5 4i?
-9
What is the conjugate of the complex number 7-4i?
To get the conjugate simply reverse the sign of the complex part. Thus conj of 7-4i is 7+4i
What is conjugate of 4i open bracket -2 -3i close bracket?
4i(-2 -3i) = 4i×-2 - 4i×-3i = -8i -12i² = -8i + 12 = 12 -8i → the conjugate is 12 + 8i
2 plus 4i - 7 plus 4i?
(2 + 4i) - (7 + 4i) = -5 2 + 4i - 7 + 4i = -5 + 8i
What is the sum of two complex conjugate number?
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.
Find the absolute value of the complex number z equals 3 plus 4i?
('|x|' = Absolute value of x) |3+4i| = √(32 + 42) = √(9+16) = √25 = 5 Thus |3+4i| = 5.
What is the multiplicative inverse of 3-4i divided by 5 plus 2i?
The multiplicative inverse of a number a is a number b such that axb=1 If we look at (3-4i)/(5+2i), we see that we can multiply that by its reciprocal and the product is one. So (5+2i)/(3-4i) is the multiplicative inverse of (3-4i)/(5+2i)
The division of two complex numbers arithematic operation requires the use of complex conjugate?
(3+2i)/(5+4i)If you multiply both sides by the conjugate of the denominator (5-4i), you get:(3+2i)(5-4i)/(5+4i)(5-4i)= (23-2i)/(25 + 16 +20i - 20i)= (23-2i)/41The denominator is now real, because the i terms cancelAs a general formula (easy to expand) this would be:(a+bi)/(c+di) = [(ac+bd) + (bc-ad)i] / (c^2 + d^2)It's a very easy method, but if you're the sort of person who loves using general formulas, there it is.
The conjugate of complex numbers 3 plus 2j is?
3-2j.
What is the complex form of 4 plus 4i plus 4 plus 6i?
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.
What is 2-4i over 4 plus 2i put in the form of a plus bi?
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.
