The discussion centers on calculating the exponent of a complex number, specifically \( z = 1 + i\sqrt{3} \) raised to the 9th power. The argument of the complex number is determined to be \( \frac{\pi}{3} \) and its magnitude is 2, leading to the expression \( z = 2e^{i\frac{\pi}{3}} \). The calculation of \( z^9 \) results in \( 512(cos(6\pi) + i sin(6\pi)) \), which simplifies to 512. However, the correct interpretation of the angle leads to \( z^9 = 512(cos(3\pi) + i sin(3\pi)) \), yielding the final result of -512.
PREREQUISITESStudents studying complex analysis, mathematicians working with complex numbers, and anyone interested in advanced mathematical concepts involving exponentiation and trigonometry.
What is [itex]\displaystyle \left(2\,e^{i\pi/3}\right)^9\,?[/itex]charmedbeauty said:Homework Statement
z= 1+i√3
find z^9
Homework Equations
The Attempt at a Solution
Arg(z) = pi/3 and |z|=2
so z= 2e^i*pi/3
so z^9 = 2^9 (cos6pi +isin 6pi)
= 512(1) =512
but the answer has negative 512?
SammyS said:What is [itex]\displaystyle \left(2\,e^{i\pi/3}\right)^9\,?[/itex]