Proof that exp z = 1 iff z is an integer multiple of 2πi

Builds on Complex exponential Integers Product of exponentials Euler's formula Power series of cosine Power series of sine Power series of the exponential function The exponential function is positive The exponential function is strictly increasing Modulus of a complex number Modulus is multiplicative Pythagorean trigonometric identity Sine is 2π-periodic Angle addition formulas

Proof

We split $z$ into real and imaginary parts and use the periods of sine and cosine.

Write $z = x + iy$ with $x, y$ real. By the product of exponentials, $e^z = e^x e^{iy}$ , where $e^w$ is the complex exponential.
At the real number $x$ , the exponential series sums to the real exponential. So $e^x$ is a positive real number.
By Euler's formula, $e^{iy} = \cos y + i \sin y$ . At real $y$ , these are the real cosine and sine. So $\cos y$ and $\sin y$ are the real and imaginary parts of $e^{iy}$ .
By the definition of the modulus and the Pythagorean identity, $\abs{e^{iy}} = \sqrt{\cos^2 y + \sin^2 y} = 1$ .
Let $k$ be an integer. Applying the periodicity of sine and cosine $\abs{k}$ times gives $\cos(2\pi k) = \cos 0 = 1$ and $\sin(2 \pi k) = \sin 0 = 0$ .
So if $z = 2 \pi i k$ , then $e^z = e^{i (2\pi k)} = \cos(2\pi k) + i \sin(2\pi k) = 1$ by step 3. This proves one direction.
Conversely, assume $e^z = 1$ . The modulus is multiplicative, so $1 = \abs{e^z} = \abs{e^x} \abs{e^{iy}} = e^x$ by steps 2 and 4.
The real exponential is strictly increasing and $e^0 = 1$ . So $e^x = 1$ forces $x = 0$ .
Then $1 = e^{iy} = \cos y + i \sin y$ . Comparing real and imaginary parts, $\cos y = 1$ and $\sin y = 0$ .
Let $k = \floor{y / (2\pi)}$ , an integer, and $t = y - 2\pi k$ . Then $0 \le t < 2\pi$ .
By periodicity, as in step 5, $\cos t = \cos y = 1$ and $\sin t = \sin y = 0$ .
Suppose that $t > 0$ . For each real $s$ , the angle addition formula gives $\sin(s + t) = \sin s \cos t + \cos s \sin t = \sin s$ .
So $t$ is a positive period of $\sin$ with $t < 2\pi$ . This contradicts the fact that $2\pi$ is the smallest positive period of $\sin$ . Hence $t = 0$ .
So $y = 2 \pi k$ and $z = x + iy = 2 \pi i k$ with $k$ an integer.

Idea

The modulus of $e^{x + iy}$ is $e^x$ and its angle is $y$ . So $e^z = 1$ forces the modulus $e^x$ to be $1$ and the angle $y$ to be a full number of turns. The minimality of the period $2\pi$ rules out any other angle.

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Proof that exp z = 1 iff z is an integer multiple of 2πi proof

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exp z = 1 iff z is an integer multiple of 2πi

Tag	`FKD07QE04C`
URI	`m2://FKD07QE04C`
Kind	proof
Writing	W0 stub
MSC	33B10
Proves	exp z = 1 iff z is an integer multiple of 2πi
Uses	Complex exponential, Integers, Product of exponentials, Euler's formula, Power series of cosine, Power series of sine, Power series of the exponential function, The exponential function is positive, The exponential function is strictly increasing, Modulus of a complex number, Modulus is multiplicative, Pythagorean trigonometric identity, Sine is 2π-periodic, Angle addition formulas
Changed	4 November 2026

Proof that exp z = 1 iff z is an integer multiple of 2πi

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