Given 4+3i

Question

A. Identify the complex conjugate

B. Multiply the conjugate pairs & write in standard form

Answer 1

 4 + 3i

Conjugate = 4 -3i

 

 

   (4 + 3i) ( 4 -3i) =  / use of distributive law twice.

 

     (4 +3i) 4 -3i ( 4 +3i)

 

      16 + 12i -12i + 9 = 25

 

        In General : ( a + bi) ( a - bi) = a^2 + b^2 / Remember this identity **

        

          In this example we have a = 4 , b=3 ;  a ^2 + b^2 = 4^2 + 3 ^2 = 16 +9 = 25

Answer 2

Starting with the addition or subtraction of 2 terms, like A + B (where A or B  can be complicated expressions including imaginary components), a conjugate is formed by reversing the sign of the second term, A-B.   In your case, it means reversing the sign of the imaginary component , 3i , of the  complex number.

Thus , the conjugate of 4 + 3i  is    4 – 3i.

Next we need the product of these two complex numbers.  Let us assume that you have had enough experience multiplying or factoring binomials to recognize this common identity. (If not then you can multiply the two binomials the hard way)

      (  X   +  Y  )  x  (X   -   Y)  =  X^2   -   Y^2.

 
Letting Y be the imaginary component,  (  4   +   3i  ) x  (  4   -   3i  )  =   4^2    -    [ 3i]^2

  =    16   -    9 ( i^ 2 )  =    16 + 9 = 25

Recalling that  i ^ 2 = -1  by definition, which changes the sign of the second component.

Since our result  is a real rather than a complex number, standard form is a bit excessive, but you could express it as --

   25  +  0i,  that is zero times i   ,  implying that there is no imaginary term.

Answer 3

A conjugate is 4-3i

B(4+3i)(4-3i) = 16 + 12i - 12i -9 i²  = 25 since i² = -1

 

Jim S

Answer 4

Conjugate of 4 + 3i is 4- 3i

Multiply conjugate pairs

 

(4+3i)(4-3i) = (4^2) - (3i)^2 = 8 - 9i^2 = 8 - 9(-1) = 8+9 = 17

Standard form = 17 +0i