Answer:
As this is a quartic equation, we should expect four roots (although some may end being repeated). Notice also that the equation is symmetrical, which suggests that there might be an element of symmetry in the answer.
First, as this equation is a quadratic in terms of z2 , we can use our formula for finding the roots of a quadratic equation to give the value(s) of z2 . [If you have a problem seeing this, use the substitution y=z2 , and you get a quadratic in y .]
z2=−4±42−4×1×16√2×1=−4±16−64√2
=−4±−48√2=−2±12−−√i=−2±23–√i
=4(−12±3√2)i
=4(cos(2kπ3)+isin(2kπ3)) , where k∈1,2
=4e2kiπ3
As this is the value of z2 , whereas we want z , we need to take the square roots of both sides
z=4–√ekiπ3+nπ , where n∈0,1
=2e(k+3n)iπ3
=2cis((k+3n)iπ3)
As we have two values of k and two values of n , this gives us four solutions.
Writing these angles in terms of degrees, we have: 60, 120, 240 and 300
Evaluating, we end up with the four solutions
z=1±3–√i∪z=−1±3–√i
You can substitute z^2 = x,
Then you get x^2+4x+16=0
Step-by-step explanation:
z=1±3–√i∪z=−1±3–√i
You can substitute z^2 = x,
Then you get x^2+4x+16=0
Then you can use classical x1,2 = (-b +- √[b^2–4ac])/2a
X1,2=(-4+-√[16–16*4])/2
After you solve that you substitute back z^2 = x
