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⎛−

3 1

2

2 2 i ⎞

= ⎜

⎛ ⎛−5π

⎞ ⎛−5π

⎞⎞

= 2 ⎜cos ⎜ ⎟ + i sin⎜ ⎟

6 6

⎝ ⎝ ⎠ ⎝ ⎠⎠

∀z∈

, arg ≡ −arg z 2

z

* 1

1 1

=

z cos θ + i sinθ

=

cos θ − i sinθ

1

z

=

( θ ) i ( θ )

= cos − + sin −

= [ 1 , −θ

]

⎡ −5π

= ⎢

2 ,

⎣ 6 ⎥

[ π ]

[ 1 , θ ]

خاصيات :

نضع :

-1

1 2

∀z∈

, arg ≡ −arg z 2

z

:

* 1

1 2 1 2

( cos θ sin θ ) ( cos α sin α )

z ⋅ z = + i + i

[ ]

arg z ⋅z ≡ arg z + arg z 2 π

[ θ ]

[ α ]

z1 = 1 ,

z2 = 1 ,

[ π ]

( cos θ cos α sin θ sin α) i ( cos θ sin α sin θ cosα)

= − + +

( θ α) i ( θ α)

= cos + + sin +

= [ 1 , θ + α ]

1 2 1 2

[ ]

arg z ⋅z ≡ arg z + arg z 2 π

*

z 2

:

z 1

إذن

لكل

برهان

و

نضع

من

:

:

:

لدينا

إذن

:

-2-

z

1

arg ≡ arg z1 − arg z2

2

z2

:

*

[ π ]

z 2

z 1

‎3‎‏-لكل

برهان

من و

:

arg

z

z

=

arg z ×

1 ⎞

1

⎜ 1

2

z2

1

= arg z1

+ arg 2

z

2

[ π ]

لدينا :

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