Complex Numbers

[Guest]

can someone explain how to do this problem. This is for complex numbers & exponential.
write given problem in form a+bi
e^((5e)^(i*pi/3)).
thank u

[shyamjayakannan]

[tex]e^{(5e)^\frac{i\pi}{3}}=e^{5^\frac{i\pi}{3}\times e^\frac{i\pi}{3}}=e^{e^{\frac{i\pi}{3}\ln5}\times e^\frac{i\pi}{3}}=e^{e^{\frac{i\pi}{3}(\ln5+1)}}=e^{\cos\left\{\frac{\pi}{3}(\ln5+1)\right\}+i\sin\left\{\frac{\pi}{3}(\ln5+1)\right\}}=e^{\cos\left\{\frac{\pi}{3}(\ln5+1)\right\}}e^{i\sin\left\{\frac{\pi}{3}(\ln5+1)\right\}}[/tex]

[tex]=e^{\cos\left\{\frac{\pi}{3}(\ln5+1)\right\}}\left(\cos{\left[\sin\left\{\frac{\pi}{3}(\ln5+1)\right\}\right]}+i\sin{\left[\sin\left\{\frac{\pi}{3}(\ln5+1)\right\}\right]}\right)[/tex]

So we have written it in the form a + ib, where [tex]\boxed{a=e^{\cos\left\{\frac{\pi}{3}(\ln5+1)\right\}}\cos{\left[\sin\left\{\frac{\pi}{3}(\ln5+1)\right\}\right]}}[/tex] and [tex]\boxed{b=e^{\cos\left\{\frac{\pi}{3}(\ln5+1)\right\}}\sin{\left[\sin\left\{\frac{\pi}{3}(\ln5+1)\right\}\right]}}[/tex]