Exercise · ZW-COMPLEX-EXPONENTIAL-009 · ★★★ Advanced
Find all solutions of \(e^{z}=1+i\) (with \(k\in\mathbb{Z}\)).
Choose one option.
Show answer
- B. \(z=\tfrac12\ln 2+i\left(\tfrac{\pi}{4}+2\pi k\right)\)
Explanation
Since \(1+i=\sqrt2\,e^{i\pi/4}\), we need \(e^{x}=\sqrt2\) (\(x=\tfrac12\ln2\)) and \(y=\tfrac{\pi}{4}+2\pi k\). Dropping \(2\pi k\) leaves a single solution; \(\ln2\) would correspond to modulus \(2\).
Review the concept: Complex exponential
The exponential \(e^{z}\) of a complex number \(z=x+iy\) is the complex number with modulus \(e^{x}\) and argument \(y\).
\[e^{z}=e^{x}(\cos y+i\sin y)\]
More practice problems: complex exponential
- ZW-COMPLEX-EXPONENTIAL-001 Compute e^iπ.
- ZW-COMPLEX-EXPONENTIAL-002 What is the modulus of e^2+3i?
- ZW-COMPLEX-EXPONENTIAL-003 Which is an argument of e^2+3i?
- ZW-COMPLEX-EXPONENTIAL-004 Compute e^1+iπ/2.
- ZW-COMPLEX-EXPONENTIAL-005 Which is a period of e^z, that is, a number p≠0 with e^z+p=e^z for all z?
- ZW-COMPLEX-EXPONENTIAL-006 Compute e^iπ/3 e^iπ/6.
- ZW-COMPLEX-EXPONENTIAL-007 How many solutions does e^z=0 have?
- ZW-COMPLEX-EXPONENTIAL-008 Find all solutions of e^z=1 (with k∈ℤ).