a-bi
a(bi)-1
not
negative bi
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Yes. By definition, the complex conjugate of a+bi is a-bi and a+bi - (a - bi)= 2bi which is imaginary (or 0)
It would be 8 minus 9i or 8-9i
The complex conjugate of 2-3i is 2+3i.
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
The multiplicative inverse of a complex number is the reciprocal of that number. To find the multiplicative inverse of 4 + i, we first need to find the conjugate of 4 + i, which is 4 - i. The product of a complex number and its conjugate is always a real number. Therefore, the multiplicative inverse of 4 + i is (4 - i) / ((4 + i)(4 - i)) = (4 - i) / (16 + 1) = (4 - i) / 17.