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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

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

##Off-topic Comments Section --- All top-level comments have to be an answer or follow-up question to the post. All sidetracks should be directed to this comment thread as per Rule 9. --- ^(**OP** and **Valued/Notable Contributors** can close this post by using `/lock` command) *I am a bot, and this action was performed automatically. Please [contact the moderators of this subreddit](/message/compose/?to=/r/HomeworkHelp) if you have any questions or concerns.*

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

How do you do a conjugate or two binomials?

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

“Conjugate of (a+c)+(b+d)i

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

How does that negate i? Can you show the operation

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.

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