Question

(1+i)^ n = 4096 where i^2 = -1
Find the value of n..?

Answer 1

$(1+i)^n= (\sqrt{2}(cos(45)+i*sin(45)))^n=2^{ \frac{n}{2} }(cos (45n)+i*sin(45n))$
$2^{n/2}(cos ( n\pi /4))+i2^{n/2}(sin (n \pi /4))=2^{12}=4096$
therefore n=24