[ 4 + j 4 √3 ] [ - 4√3 + j 4 ]
- 16√3 + j 16 - j 48 - 16 √3
- 32 √3 - j 32 = (-32) [ √3 + 1 ]
-16 - 16 Sqrt[3] +i (16 - 16 Sqrt[3])
z₁ = 4 + 4√3 i = 8 (cos π/3 + i sin π/3)
z₂ = −4√3 + 4i = 8 (cos 5π/6 + i sin 5π/6)
In standard form:
z₁ z₂
= (4 + 4√3 i) (−4√3 + 4i)
= 4 (−4√3 + 4i) + 4√3 i (−4√3 + 4i)
= −16√3 + 16i − 48i + 16√3 i²
= −16√3 + 16i − 48i − 16√3
= −32√3 − 32i
In trig form:
= 8 (cos π/3 + i sin π/3) * 8 (cos 5π/6 + i sin 5π/6)
= (8*8) (cos(π/3+5π/6) + i sin(π/3+5π/6))
= 64 (cos 7π/6 + i sin π/6)
= 64 (−√3/2 − 1/2 i)
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[ 4 + j 4 √3 ] [ - 4√3 + j 4 ]
- 16√3 + j 16 - j 48 - 16 √3
- 32 √3 - j 32 = (-32) [ √3 + 1 ]
-16 - 16 Sqrt[3] +i (16 - 16 Sqrt[3])
z₁ = 4 + 4√3 i = 8 (cos π/3 + i sin π/3)
z₂ = −4√3 + 4i = 8 (cos 5π/6 + i sin 5π/6)
In standard form:
z₁ z₂
= (4 + 4√3 i) (−4√3 + 4i)
= 4 (−4√3 + 4i) + 4√3 i (−4√3 + 4i)
= −16√3 + 16i − 48i + 16√3 i²
= −16√3 + 16i − 48i − 16√3
= −32√3 − 32i
In trig form:
z₁ z₂
= 8 (cos π/3 + i sin π/3) * 8 (cos 5π/6 + i sin 5π/6)
= (8*8) (cos(π/3+5π/6) + i sin(π/3+5π/6))
= 64 (cos 7π/6 + i sin π/6)
= 64 (−√3/2 − 1/2 i)
= −32√3 − 32i