simplify theses complex numbers to its smallest form (7+3i)(-7+5i)(-8-4i)

Question

can i get help to solve theses complex numbers.

Answer 1

Multiplication in complex numbers (a+bi)(c+di) = (ac-bd)+(bc+ad)i

Given expression is(7+3i)(7-3i)(-8-4i)

Now (7+3i)(7-3i) = 7^2-(3i)^2

= 49-9i^2

= 49-9*(-1)

= 49+9

= 58

Now (7+3i)(7-3i)(-8-4i) = 58(-8-4i)

= -464-232i

                                                                                                                                                                                                       

Comments

The complex number expression is (7 + 3i)(- 7 + 5i)(- 8 - 4i).

Answer 2

Multiplication in complex numbers (a+bi)(c+di) = (ac-bd)+(bc+ad)i

Given expression is(7+3i)(-7+5i)(-8-4i)

Now (7+3i)(-7+5i)

a = 7,b = 3, c = -7, d = 5

=[(7*-7)-(3*5)]+[(3*-7)+(7*5)]

= (-49-15)+[( -21+35)i]

= -64+14i

Now (7+3i)(-7+5i)(-8-4i) = (-64+14i)(-8-4i)

a = -64, b = 14,c = -8, d = -4

= [(-64*-8)-(14*-4)]+[(14*-8)+(-64*-4)i]

= (512-56)+(-112+256)i

= 456+144i

Answer 3

Multiplication of complex numbers : (a + bi)(c + di)= (ac- bd) + (bc + ad)i.

The expression is (7 + 3i)(- 7 + 5i)(- 8 - 4i).

 

First find (7 + 3i)(- 7 + 5i).

Compare it with (a + bi)(c + di).

a = 7, b = 3, c = - 7, and d = 5.

(7 + 3i)(- 7 + 5i) = [ (7 * - 7) - (3 * 5) ] + [ (3 * - 7) + (7 * 5) ]i

= [ - 49 - 15 ] + [ - 21 + 35 ]i

= (- 64 + 14i).

 

(7 + 3i)(- 7 + 5i)(- 8 - 4i) = (- 64 + 14i)(- 8 - 4i).

Compare (- 64 + 14i)(- 8 - 4i) with (a + bi)(c + di).

a = - 64, b = 14, c = - 8, and d = - 4.

(- 64 + 14i)(- 8 - 4i) = [ (- 64 * - 8) - (14 * - 4) ] + [ (14 * - 8) + (- 64 * - 4) ]i

= [ (512) + 56 ] + [ - 112 +256 ]i

= 568 +144i.

 

∴ (7 + 3i)(- 7 + 5i)(- 8 - 4i) = 568 +144i.