SOLUTION: Express the equation z^4=−2 + 2√3i in polar form. Hence solve the equation, expressing your answers in polar form

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Question 681027: Express the equation z^4=−2 + 2√3i in polar form. Hence solve the equation,
expressing your answers in polar form

Answer by Edwin McCravy(20055) About Me  (Show Source):
You can put this solution on YOUR website!
z%5E4%22%22=%22%22-2+%2B+2sqrt%283%29i

matrix%282%2C1%2C%22%22%2Cz%29matrix%282%2C1%2C%22%22%2C%22%22=%22%22%29matrix%282%2C1%2C%22%22%2C%282+%2B+2sqrt%283%29i%29%5E%281%2F4%29%29

x+=+-2, y+=2sqrt%283%29, r=sqrt%28x%5E2%2By%5E2%29=sqrt%28%28-2%29%5E2%2B%282sqrt%283%29%29%5E2%29 = sqrt%284%2B4%2A3%29 = sqrt%284%2B12%29 = sqrt%2816%29 = 4
 
tan%28theta%29%22%22=%22%22y%2Fx%22%22=%22%22%282sqrt%283%29%29%2F%28-2%29%22%22=%22%22-sqrt%283%29.

The reference angle for theta is 60° and since x is - and y is +,
theta is in the 2nd quadrant, so we subtract 60° from 180° and get
120°, which has the same trig functions as 120° + 360°·k, so we take:

theta%22%22=%22%22%22120%B0%22%2B%22360%B0%22k.

Now we have

matrix%282%2C1%2C%22%22%2Cz%29matrix%282%2C1%2C%22%22%2C%22%22=%22%22%29matrix%282%2C1%2C%22%22%2C%282+%2B+2sqrt%283%29i%29%5E%281%2F4%29%29matrix%282%2C1%2C%22%22%2C%22%22=%22%22%29matrix%282%2C1%2C%22%22%2C%28r%28cos%28theta%29%2Bi%2Asin%28theta%29%29%29%5E%281%2F4%29%29matrix%282%2C1%2C%22%22%2C%22%22=%22%22%29

Now we use the formula: %28r%28cos%28theta%29%2Bi%2Asin%28theta%29%29%29%5En%22%22=%22%22r%5En%2A%28cos%28n%2Atheta%29%2Bi%2Asin%28n%2Atheta%29%29

matrix%282%2C1%2C%22%22%2Cz%29matrix%282%2C1%2C%22%22%2C%22%22=%22%22%29matrix%282%2C1%2C%22%22%2C%22%22=%22%22%29



matrix%282%2C1%2C%22%22%2C%22%22=%22%22%29
sqrt%282%29%28cos%28%2230%B0%22%2B%2290%22k%29%2Bi%2Asin%28%2230%B0%22%2B%2290%22k%29%29

Now we let k = 0,1,2,and 3 to get the 4 4th roots, the 4 possible
answers for z:

For k = 0,

%22%22=%22%22sqrt%282%29%28cos%28%2230%B0%22%29%2Bi%2Asin%28%2230%B0%22%29%29

For k = 1,

%22%22=%22%22sqrt%282%29%28cos%28%22120%B0%22%29%2Bi%2Asin%28%22120%B0%22%29%29

For k = 2,

%22%22=%22%22sqrt%282%29%28cos%28%22210%B0%22%29%2Bi%2Asin%28%22210%B0%22%29%29

For k = 3,

%22%22=%22%22sqrt%282%29%28cos%28%22300%B0%22%29%2Bi%2Asin%28%22300%B0%22%29%29

Those are the four values for z.

Edwin