-5+4i
HNO2 conjugate acid = one more hydrogen conjugate base = one less hydrogen
The conjugate base and conjugate acid for HS04 is: Conjugate acid is H2SO4 Conjugate base is SO42
NH2- is the conjugate base of ammonia.
The conjugate acid of F- is HF.
Acid + base conjugate base + conjugate acid
When finding the conjugate of a binomial, you just reverse the sign. So the conjugate of 3+4i is 3-4i.
-6-4i.
To get the conjugate simply reverse the sign of the complex part. Thus conj of 7-4i is 7+4i
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.
4i(-2 -3i) = 4i×-2 - 4i×-3i = -8i -12i² = -8i + 12 = 12 -8i → the conjugate is 12 + 8i
(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 + 4i) - (7 + 4i) = -5 2 + 4i - 7 + 4i = -5 + 8i
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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)
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('|x|' = Absolute value of x) |3+4i| = √(32 + 42) = √(9+16) = √25 = 5 Thus |3+4i| = 5.
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