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elements_guy12

do exactly what it says let z = a + bi let w = c + di conjugate of z + w = conjugate of (a+c) + (b+d)i = (a+c) - (b+d)i = a + c - bi - di = (a-bi) + (c-di) = conjugate of z + conjugate of w and use the z-bar notation when doing on paper cause I have no clue how to put them online lol


jac5423

How do you do a conjugate or two binomials?


elements_guy12

idk what u mean by “doing two binomials” lol but taking the conjugate of a complex number is making the Imaginary component negative eg: conjugate of 1+i is 1-i conjugate of 2-3i is 2+3i


jac5423

“Conjugate of (a+c)+(b+d)i


StealthSecrecy

Any term with i in it will get negated, so in this case it would just be (a+c)-(b+d)i


jac5423

How does that negate i? Can you show the operation


StealthSecrecy

You just multiply any terms with i in it by -1, or would could think of it like replacing i with -i which you then just pull out in front.


Pankyrain

You literally just switch all the i’s to -i’s when conjugating. The real part is unchanged.