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as a2+b2=(a+ib)(a-ib), b2+2=[b - i sqrt(2)] [b + i sqrt(2)]
It already is in the form a+ib. a = root 7, the b=1 (i.e 1 x i ).
No
If you have a complex function in the form "a+ib", the (complex) conjugate is "a-ib". "Conjugate" is usually a function that the original function must be multiplied by to achieve a real number.
Numbers of the form a + ib (where i = -11/2; a and b are real numbers) are called complex numbers.For any two complex numbers (a+ib) and (c+id):Addition:(a+ib) + (c+id) = (a+c) + i(b+d)So, -8i-7i = (-8-7)i = -15i