Calculate: (1+3i)^9 How do I solve this? I have tried without any success. Thanks!
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Originally Posted by sf1903 Calculate: (1+3i)^9 $\displaystyle 1 + i\sqrt 3 = 2\cos \left( {\frac{\pi } {3}} \right) + i2\sin \left( {\frac{\pi } {3}} \right)$ $\displaystyle 2^9=512$ and $\displaystyle 9\left(\frac{\pi}{3}\right)=3\pi$
Hint : $\displaystyle 2e^{i\pi/3}=1+\sqrt{3}i$.
Originally Posted by Plato $\displaystyle 1 + i\sqrt 3 = 2\cos \left( {\frac{\pi } {3}} \right) + i2\sin \left( {\frac{\pi } {3}} \right)$ $\displaystyle 2^9=512$ and $\displaystyle 9\left(\frac{\pi}{3}\right)=3\pi$ If im correct the answer should be -512 512*Cos(3π)+i*512*Sin(3π) = -512 + 0 = -512 Or am i out in the blue?
Yes! $\displaystyle (2e^{i\pi/3})^9=512e^{3i\pi}=-512$.
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