Please help me with my Trigonometry?

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( 1 )

(1 + i )⁵

= ( √2 )⁵ [ cos(5 * ¼π ) + isin(5 * ¼π ) ]

= 4√2 ( -½√2 - ½i√2 )

= -4 - 4i

= -4 (1 + i )

━━━━━

( 2 )

Assuming (-√3 + i)⁵

arctan (1/(-√3) ) = -⅙π = 7π/6

(-√3 + i)⁵

= ( 2 )⁵ [ cos(5 * 7π/6 ) + isin(5 * 7π/6 ) ]

= 32 ( ½√3 - ½i )

= 16( √3 - i )

━━━━━━

1.

(1 + i)^5

= -4 - 4 i

2.

(-√3 + i)^5

= 27.7128129211020366964391414640939578710848840609660900488... +

16 i

(1 +i)^5 =

Use Pascal's triangle

1^5 + 5*1^4*i + 10*1^3*i^2 + 10*1^2*i^3 + 5*1*i^4 + i^5 =

1 + 5i -10 - 10i +5 + i =

-4 - 4i

(-√3 + i)^5 =

(-√3)^5 + 5*(-√3)^4*i + 10*(-√3)^3*i^2 + 10*(-√3)^2*i^3 + 5*(-√3)*i^4 + i^5 =

-9√3 + 45i +30√3 -30i -5√3 -i =

16√3 +14i

(1 + i)^5 = (√2 e^(i*π/4))^5 = 4√2 e^(i*5π/4) = -4 - 4i

or

(1 + i)^5 = 1^5 + 5*1^4*i^1 + 10*1^3*i^2 + 10*1^2*i^3 + 5*1^1*i^4 + i^5 = -4 - 4i

Use either of these methods again to answer question 2.