1) i^(2-i),2) (-1-i)^(1+i)

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摘要 z = e^(iθ) = cosθ + isinθ = x + iyzⁿ = e^(inθ) = cos(nθ) + isin(nθ) = (x + iy)ⁿarg(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)]

咨询记录 · 回答于2022-09-23

1) i^(2-i),2) (-1-i)^(1+i)

亲，您这个题目乱码了，能拍个照片发过来吗？这样没法解题哦，亲

好的，亲

好了吗

i的2-i次方 = i的平方 / i 的 i 次方分子中： i的平方 = -1i=e^(2*k*π*i+pi/2 *i)i^i=e^(i*(2*k*π*i+π/2 *i))=e^(-2*k*π-π/2) i的2-i次方 = - e^(2*k*π+π/2)

第一问，如上

z = e^(iθ) = cosθ + isinθ = x + iyzⁿ = e^(inθ) = cos(nθ) + isin(nθ) = (x + iy)ⁿarg(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)]

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