Using De Moivre's Theorem, the complex number (2sqrt2-2isqrt2)^6 in rectangular form is -4096. The modulus of the given complex number is 4, and the argument is -pi/4. Raising it to the 6th power and simplifying gives a real number -4096.
To find (2sqrt2-2isqrt2)^6 in rectangular form using De Moivre's Theorem, we first recognize that the complex number 2sqrt2-2isqrt2 can be written in polar form.
The modulus (r) of this number is sqrt((2sqrt2)^2 + (2sqrt2)^2) = 4, and its argument (theta) is tan-1(-1) = -pi/4 or 7pi/4 when expressed in radians. De Moivre's Theorem states that (r(cos(theta) + i*sin(theta)))^n = r^n * (cos(n*theta) + i*sin(n*theta)).
Applying this to our given complex number:
(4(cos(7pi/4) + i*sin(7pi/4)))^6 = 4^6 * (cos(6 * 7pi/4) + i*sin(6 * 7pi/4))
Considering that sin and cos have a period of 2pi, we can reduce the angle by subtracting multiples of 2pi. So, cos(6 * 7pi/4) becomes cos(6pi/2) which is cos(3pi) and sin(6 * 7pi/4) becomes sin(6pi/2) which is sin(3pi).
The final step is to raise r to the 6th power and apply the cosine and sine of the resulting angle
4^6 * (cos(3pi) + i*sin(3pi))
= 4096 * (-1 + 0i)
= -4096 (since sin(3pi) = 0 and cos(3pi) = -1).
So in rectangular form, (2sqrt2-2isqrt2)^6 equals -4096.