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What is the conjugate of -5 4i?

Updated: 9/21/2023
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11y ago

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-9

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Ethel Hauck

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3y ago
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12y ago

-5+4i

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Q: What is the conjugate of -5 4i?
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What is the conjugate of 3 plus 4i?

When finding the conjugate of a binomial, you just reverse the sign. So the conjugate of 3+4i is 3-4i.


What is the conjugate of -6 plus 4i?

-6-4i.


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 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.


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


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.


2 plus 4i - 7 plus 4i?

(2 + 4i) - (7 + 4i) = -5 2 + 4i - 7 + 4i = -5 + 8i


What is multiplies to 20 but adds to 4?

24


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)


What two numbers multiply to 20 but add together to 4?

24


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 equation for whose roots are 4i -3i and 3i?

The equation is:x3 - 4x2i - 9xi2+ 36i3 = 0Note: The ' i ' must be an ordinary constant, and can't be sqrt(-1).If it were sqrt(-1), then -4i would also be a root of the equation.Imaginary or complex roots always occur in conjugate pairs.If -4i is also a root of the equation and the questioner just forgot to include it,then ' i ' can be sqrt(-1), and the equation can bex4 + 25x2 + 144 = 0