Question

Show that if eᶻ is real, then Im z = nπ (n = 0, ±1, ±2, . . .).

Answer 1

If
`\(e^z\)` is real, then
`\(Im(z) = n\pi\)` where
`\(n\)` is an integer (i.e.,
`\(n = 0, \pm1, \pm2, \ldots\)`).

Step-by-step explanation:

When dealing with complex numbers,
`\(z\)` can be expressed as
`\(z = x + yi\)`, where
`\(x\)` and
`\(y\)` are real numbers, and
`\(i\)` is the imaginary unit
`(\(i^2 = -1\))`. The real exponential function
`\(e^z\)` is defined as
`\(e^z = e^(x + yi) = e^x \cdot e^(iy)\)`.

For
`\(e^z\)` to be real, the imaginary part
`(\(Im(z)\)) of \(z\)` must satisfy
`\(e^(iy)\)` being real. The imaginary exponential function
`\(e^(iy)\)` can be expressed using Euler's formula as
`\(e^(iy) = \cos(y) + i \sin(y)\)`. For
`\(e^(iy)\)` to be real,
`\(\sin(y)\)` must be zero, which occurs when
`\(y = n\pi\),` where
`\(n\)` is an integer.

Therefore,
`\(Im(z) = y = n\pi\)` when
`\(e^z\)` is real.

In conclusion, if
`\(e^z\)` is real, the imaginary part
`\(Im(z)\)` takes values of
`\(n\pi\)`, where
`\(n\)` is an integer. This is a consequence of the trigonometric properties of the imaginary exponential function, ensuring that the imaginary part results in multiples of
`\(\pi\)` for
`\(e^z\)` to be a real number.