Question

(a +ib)-(c+ id) =



please answer this​

Answer 1

Answer:

a + ib)/(c + id) (when you divide by a complex number you need to multiply top and bottom by its conjugate, in this case c - id)

(a + ib)/(c + id) * (c - id)/(c - id)

((a + ib)(c - id))/((c + id)(c - id)) (use FOIL on top and bottom)

(ac - iad + ibc - (i^2)bd)/(c^2 - icd + icd - (i^2)d^2)

(ac - iad + ibc - (-1)bd)/(c^2 - icd + icd - (-1)d^2)

(ac + bd - iad + ibc)/(c^2 + d^2)

(ac + bd + i(bc - ad))/(c^2 + d^2)

ac/(c^2 + d^2) + bd/(c^2 + d^2) + i((bc - ad)/(c^2 + d^2))

A = (ac + bd)/(c^2 + d^2)

B = (bc - ad)/(c^2 + d^2)

----------

(a - ib)/(c - id) (conjugate of c - id is c + id)

(a - ib)/(c - id) * (c + id)/(c + id)

((a - ib)(c + id))/((c - id)(c + id)) (use FOIL on top and bottom)

(ac + iad - ibc - (i^2)bd)/(c^2 + icd - icd - (i^2)d^2)

(ac + iad - ibc - (-1)bd)/(c^2 - (-1)d^2)

 

(ac + bd - i(bc - ad))/(c^2 + d^2)

ac/(c^2 + d^2) + bd/(c^2 + d^2) - i((bc - ad)/(c^2 + d^2))

A = (ac + bd)/(c^2 + d^2)

B = (bc - ad)/(c^2 + d^2)

----------

1. (a+ib)/(c+id) = ac/(c^2 + d^2) + bd/(c^2 + d^2) + i((bc - ad)/(c^2 + d^2))

(a+ib)/(c+id)=A+iB

A = (ac + bd)/(c^2 + d^2)

B = (bc - ad)/(c^2 + d^2)

2. (a-ib)/(c-id) = ac/(c^2 + d^2) + bd/(c^2 + d^2) - i((bc - ad)/(c^2 + d^2))

(a-ib)/(c-id)=A-iB

A = (ac + bd)/(c^2 + d^2)

B = (bc - ad)/(c^2 + d^2)

----------

proven

Step-by-step explanation:

Answer 2

Answer:

(a-c) +i(b-d) will be answer.