請幫忙計算以下數學算式

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1) Find the square root of z = 3-4i.
[Hint: let x + yi = √z ]

Let the square root of z = 3-4i be x + yi where x and y are real,
then (x+yi)² = z = 3-4i
x²-y² +2xyi = 3 - 4i

{ x²-y² = 3
{ 2xy = - 4 => y = -2/x

Therefore, x²-(-2/x)² = 3

x² - 4/x² = 3
x⁴ - 3x² - 4 = 0
(x² - 4)(x² + 1) = 0
x² = 4 or x² = -1 (rejected because x is real)
x = 2 or -2
If x = 2, y = -1
If x = -2, y = 1
Therefore, the square root of z = 3 - 4i can either be 2 - i or -2 + i

2) Covert the following complex numbers to polar form
a) - i

-i = 0 + (-1) i = cos(3π/2) + sin(3π/2)i

b) -4+3i

Consider √[4²+3²] = 5.
Then,
-4+3i
= 5 [ -4/5 + 3/5 i]
= 5 [ cos(θ) + sin(θ) i]

where θ = π - arcsin(3/5) = 2.498 = 143.13°

2013-10-04 02:21:38 補充：
Mr. Kwok 你好～
你說得對～
剛剛就打算關機了～

謝謝關心～
^_^

2013-10-05 23:04:22 補充：
郭老師問我的時候是星期五的凌晨，他應該是指星期五那一天。

...............

2013-10-05 20:22:33 補充：
Still work on Saturday??? :O

Masterijk 晚上好，如此晚了還未睡，不用上班嗎？