Question

求复数（1+i）^i的实部和虚部大概是个什么思路

Answer 1

(1+i)^i=e^(i*ln(1+i))
ln(1+i)=ln(1/√2+1/√2i)+ln(√2)
=
(π/大锋4)i+1/2*ln(2)
i*ln(1+i)
=
-π/4
+1/2*ln(2)
i
e^
(-π/4
+1/2*ln(2)
i
)
=
e^(-π/悉嫌4)
*
e^(1/2*ln(2)
i
)
=
e^(-π/4)
*
(
cos(ln(2)/2)
+
i
*
sin(ln(2)/2)
)
因此：
(1+i)^i
的实部
e^(-π/4)
*
cos(ln(2)/2)
=
0.428829006
(1+i)^i
的虚部
e^(-π/4)
*
sin(ln(2)/2)
=
0.154871752
希望有帮助，不清楚请追问，有用请滚陆晌采纳
o(∩_∩)o

Answer 2

(1+i)^i
=e^[iLn(1+i)]
=e^{i[ln|1+i|+iarg(1+i)+i2kπ]}
=e^{i[ln√脊稿锋樱晌2+iπ/4+i2kπ]}
=e^(iln√敬州2-π/4-2kπ)