- Square Root of a Complex Number

LESSON 1

Find \(\sqrt{15+8i}\)

SOLUTION
We assume that the square root of a complex number is a complex number
\(\sqrt{15+8i}=a+bi\)
  • Square both sides and simplify
\(15+8i=(a+bi)^2\)
\(15+8i=a^2-b^2+2abi\)
  • Equate real parts
\(a^2-b^2=15\) (1)
  • Equate imaginary parts
\(2ab=8\)
\(\displaystyle a={4\over b}\) (2)
  • Substitute (2) into (1) and solve for \(a\) and \(b\)
\(\displaystyle \left({4\over b}\right)^2-b^2=15\)
\(\displaystyle {16\over b^2} -b^2=15\)
\(b^4+15b^2-16=0\)
\((b^2+16)(b^2-1)=0\)
\(b^2=-16\) Invalid since \(b\) is real
\(b^2=1\)
\(b=±1\)
\(a=±4\)
  • Substitute values for a and b into a+bi
\(\sqrt{15+8i}=4+i\) or \(-4-i\)
\(=±(4+i)\)
Thus, we see that a complex number has 2 square roots, which are complex numbers.