如何证明复数乘法对应复平面旋转？如何推导复指数表达式？
Great question—let's break this down step by step, since understanding how complex numbers act as rotations ties together algebra, geometry, and even calculus a bit.
Let's start with the concrete example you gave, because it's a perfect way to build intuition first. Take a first-quadrant complex number z₀ = a + bi (yes, talking about quadrants in the complex plane makes total sense—we just map the real part to the x-axis and imaginary part to the y-axis, so it's just a 2D coordinate system).
When you multiply z₀ by i:
z₁ = i * z₀ = i(a + bi) = ai + b*i² = -b + ai
If you plot this, z₀ was at (a, b); z₁ is at (-b, a). That's a 90-degree counterclockwise rotation around the origin! Multiply by i again:
z₂ = i * z₁ = i(-b + ai) = -bi + a*i² = -a - bi
Now we're at (-a, -b)—another 90-degree rotation, so 180 total from the original. Do it a third time, you get b - ai (270 degrees), and a fourth time brings you back to a + bi. This pattern alone tells us multiplying by i is a rotation, but we can generalize this to any complex number, not just i.
The connection between complex numbers and rotations really clicks when we derive Euler's Formula, which links the exponential function to trigonometry. Let's use Taylor series expansions, since they're a straightforward way to bridge these concepts.
First, recall the Taylor series for eˣ, sin(x), and cos(x) around 0:
eˣ = 1 + x + x²/2! + x³/3! + x⁴/4! + x⁵/5! + ...cos(x) = 1 - x²/2! + x⁴/4! - x⁶/6! + ...sin(x) = x - x³/3! + x⁵/5! - x⁷/7! + ...
Now substitute x = iθ into the exponential series (where i is the imaginary unit, i² = -1):
e^(iθ) = 1 + iθ + (iθ)²/2! + (iθ)³/3! + (iθ)⁴/4! + (iθ)⁵/5! + ...
Let's expand each term using powers of i:
(iθ)² = i²θ² = -θ²(iθ)³ = i³θ³ = -iθ³(iθ)⁴ = i⁴θ⁴ = θ⁴(iθ)⁵ = i⁵θ⁵ = iθ⁵- And so on, since
i^ncycles every 4:1, i, -1, -i, 1, i, ...
Now group the real and imaginary terms separately:
e^(iθ) = [1 - θ²/2! + θ⁴/4! - ...] + i[θ - θ³/3! + θ⁵/5! - ...]
Hey, those are exactly the Taylor series for cos(θ) and sin(θ)! So we get:
e^(iθ) = cosθ + i sinθ
This is Euler's Formula, and it's the key to seeing why complex exponents relate to rotation: e^(iθ) is a complex number on the unit circle (since its magnitude is √(cos²θ + sin²θ) = 1) at an angle θ from the positive real axis.
Now let's formalize this for any complex number. Any complex number z can be written in polar form (which is perfect for rotation):
z = r * (cosθ + i sinθ)
Where r is the magnitude (distance from the origin, √(a² + b²) if z = a + bi) and θ is the angle from the positive real axis.
Suppose we multiply z by another complex number w (also in polar form):
w = s * (cosφ + i sinφ)
Let's compute the product z * w:
z*w = r*s * (cosθ + i sinθ)(cosφ + i sinφ)
Expand the trigonometric part using the cosine and sine addition formulas:
(cosθ + i sinθ)(cosφ + i sinφ) = cosθcosφ - sinθsinφ + i(sinθcosφ + cosθsinφ) = cos(θ + φ) + i sin(θ + φ)
So putting it all together:
z*w = r*s * [cos(θ + φ) + i sin(θ + φ)]
What does this mean? The magnitude of the product is r*s (we scale the original length by s), and the angle is θ + φ—we've rotated the original complex number z by φ radians counterclockwise around the origin!
If we want a pure rotation (no scaling), we just take s = 1—so w is a unit complex number, w = cosφ + i sinφ = e^(iφ). Your original example with multiplying by i is just a special case here: i = cos(π/2) + i sin(π/2) = e^(iπ/2), so multiplying by i is a 90-degree (π/2 radian) rotation, which matches exactly what you saw with z₀ → z₁ → z₂.
内容的提问来源于stack exchange，提问作者user525966
