复变函数,1-i用三角函数如何表示?1-i的i次方应如何求解?网上有一些输入规则有点容易混淆,恳请大神们能手写之后拍照上传,

Question

复变函数,1-i用三角函数如何表示?1-i的i次方应如何求解?
网上有一些输入规则有点容易混淆,恳请大神们能手写之后拍照上传,

Answer 1

z = e^(iθ) = cosθ + isinθ = x + iy
zⁿ = e^(inθ) = cos(nθ) + isin(nθ) = (x + iy)ⁿ
arg(z) = arctan(y/x)
|z| = √(x² + y²)
∵arg(z) = - π/4
|z| = √(1² + (- 1)²) = √2
∴1 - i
= √2e^(- iπ/4)
= √2[cos(- π/4) + isin(- π/4)]
= √2[cos(π/4) - isin(π/4)]
∵arg(z) = - π/4
|z|^i = (1² + 1²)^(i/2) = 2^(i/2)
∴(1 - i)^i
= 2^(i/2) • e^(i • i • - π/4)
= 2^(i/2) • e^(π/4)
= 2^(i/2)[cos(π/4) + isin(π/4)]
或
= e^(π/4) • [cos(1/2 • ln2) + isin(1/2 • ln2)]
∵arg(z) = π/4
|z| = √(1² + 1²) = √2
∴1 + i
= √2e^(iπ/4)
= √2[cos(π/4) + isin(π/4)]
∵arg(z) = π/4
|z|^(1/4) = (1² + 1²)^(1/2 • 1/4) = 2^(1/8)
∴(1 + i)^(1/4)
= 2^(1/8) • e^(i • 1/4 • π/4)
= 2^(1/8) • e^(iπ/16)
= 2^(1/8) • [cos(π/16) + isin(π/16)]