Exercise · ZW-DE-MOIVRE-FORMULA-001 · ★☆☆ Basic
Compute \((1+i)^8\).
Choose one option.
Show answer
- C. \(16\)
Explanation
\(1+i=\sqrt2\,(\cos\frac\pi4+i\sin\frac\pi4)\), so \((1+i)^8=(\sqrt2)^8(\cos2\pi+i\sin2\pi)=16\). A common slip is using \(r=2\), which would give 256.
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-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-006 For n=-1, (cos θ+isin θ)^-1 equals:
- 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: