z= 10 (cos(56π)+isin(56π))in polar form

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
${\large z=10\exp(56\pi\bf{i})}$ 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))

${\large z=10\exp(56\pi\bf{i})}$ so $z^3=10^3\exp(168\pi\bf{i})$.
BUT is it true that $z=10~?$
If it really were " $56\pi$" then, yes, it would be 10.
With $5\pi/6$, $z^2= 10^3(exp(3*(5/6)\pi i)= 1000 exp((5/2)\pi i)= 1000 exp((1/2)\pi)= 1000(cos(\pi/2)+ i sin(\pi/2))= 1000i$.