Exercise · ZW-DE-MOIVRE-FORMULA-006 · ★★☆ Intermediate
For \(n=-1\), \(\left(\cos\theta+i\sin\theta\right)^{-1}\) equals:
Choose one option.
Show answer
- B. \(\cos\theta-i\sin\theta\)
Explanation
The formula with \(n=-1\) gives \(\cos(-\theta)+i\sin(-\theta)=\cos\theta-i\sin\theta\), the conjugate, which is the inverse of a unit-modulus complex number.
Review the concept: De Moivre's formula
De Moivre's formula gives the integer power \(n\) of a complex number \(z\) of modulus \(r\) and argument \(\theta\): it raises the modulus to the \(n\)-th power and multiplies the argument by \(n\).
\[z^n=r^n\left(\cos n\theta+i\sin n\theta\right)\]
More practice problems: de moivre's formula
- ZW-DE-MOIVRE-FORMULA-001 Compute (1+i)^8.
- ZW-DE-MOIVRE-FORMULA-002 Compute (cos π6+isin π6)^6.
- ZW-DE-MOIVRE-FORMULA-003 Use the formula to find (1-i)^4.
- ZW-DE-MOIVRE-FORMULA-004 What is the modulus of (√3+i)^5?
- ZW-DE-MOIVRE-FORMULA-005 Compute (1+i√3)^3.
- ZW-DE-MOIVRE-FORMULA-007 Expanding (cos θ+isin θ)^3 and equating real parts, what expression results for cos 3θ?
- ZW-DE-MOIVRE-FORMULA-008 Compute (√3-i)^6.
- ZW-DE-MOIVRE-FORMULA-009 If n is an integer and you replace θ by θ+2π in r^n(cos nθ+isin nθ), the value: