z= 10 (cos(56 π)+isin(56 π))in polar form I'm not sure how to put z^3 in Cartesian form ..please help
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56pi (or any even multiple of pi) is coterminal with 2pi. So, cos (56pi) = cos (2pi) = 1. Similarly sin(56pi) = sin(2pi) = 0 So z = 10(1 + 0i) =10. Now z^3 is easy!! (unless you meant 5pi/6 instead of 56pi)
Originally Posted by bee77 z= 10 (cos(56 π)+isin(56 π))in polar form I'm not sure how to put z^3 in Cartesian form ..please help ${\large z=10\exp(56\pi\bf{i})}$ so $z^3=10^3\exp(168\pi\bf{i})$. BUT is it true that $z=10~?$
Oh sorry it should have been 5/6 PI i with both the cos and the sine ....
terribly sorry ,yes I meant 5pi/6
Originally Posted by Plato ${\large z=10\exp(56\pi\bf{i})}$ so $z^3=10^3\exp(168\pi\bf{i})$. BUT is it true that $z=10~?$ Yes because exp(168pi i) = 1
Recall that if z = r(cos (A) + i sin (A)) then z^3 = (r^3)(cos(3A)+i sin(3A))
Last edited by Debsta; Mar 20th 2017 at 09:30 PM.
Thanks a heap ,I'm pretty rusty on trig as it's been a very long time...really appreciate the help
Originally Posted by Plato ${\large z=10\exp(56\pi\bf{i})}$ so $z^3=10^3\exp(168\pi\bf{i})$. BUT is it true that $z=10~?$ If it really were "$\displaystyle 56\pi$" then, yes, it would be 10. With $\displaystyle 5\pi/6$, $\displaystyle z^2= 10^3(exp(3*(5/6)\pi i)= 1000 exp((5/2)\pi i)= 1000 exp((1/2)\pi)= 1000(cos(\pi/2)+ i sin(\pi/2))= 1000i$.