Question 1086747: Perform the indicated operations using trigonometric form. Please leave your answer in trigonometric form. (1+i)^4
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website! z = 1+i = 1+1*i is in the form a+b*i where a = 1 and b = 1
r = sqrt(a^2+b^2)
r = sqrt(1^2+1^2)
r=sqrt(2)
theta = arctan(b/a)
theta = arctan(1/1)
theta = arctan(1)
theta = pi/4 radians (45 degrees)
If z = 1+i, then z = sqrt(2)*(cos(pi/4) + i*sin(pi/4)) is the trig form of the given complex number.
Recall that z = r*(cos(theta)+i*sin(theta)) is the general trig form.
De Moivre's Theorem is the idea of raising a trig complex number to some integer power n
z = r*(cos(theta)+i*sin(theta))
z^n = [r*(cos(theta)+i*sin(theta))]^n
z^n = r^n*(cos(n*theta)+i*sin(n*theta))
Using De Moivre's Theorem, we can say
z^n = r^n*(cos(n*theta)+i*sin(n*theta))
z^4 = r^4*(cos(4*theta)+i*sin(4*theta))
z^4 = (sqrt(2))^4*(cos(4*pi/4)+i*sin(4*pi/4))
z^4 = 4*(cos(pi)+i*sin(pi))
z^4 = 4*(-1+i*0)
z^4 = 4*(-1+0)
z^4 = 4*(-1)
z^4 = -4
Earlier we defined z = 1+i and we just found out z^4 = -4
So in conclusion, (1+i)^4 = -4
If you wish to leave the answer in trig form, then writing it as 4*(cos(pi)+i*sin(pi)) should be fine (though it won't be fully simplified)
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