Square Root of Complex Number in Cartesian Form/Examples/8 + 4 root 5 i

Example of Square Root of Complex Number in Cartesian Form

$\sqrt {8 + 4 \sqrt 5 i} = \pm \paren {\sqrt {10} + \sqrt 2 i}$

Proof

\(\ds \paren {x + i y}^2\)

\(=\)

\(\ds 8 + 4 \sqrt 5 i\)

\(\ds \leadsto \ \ \)

\(\ds x^2\)

\(=\)

\(\ds \dfrac {8 + \sqrt {8^2 + \paren {4 \sqrt 5}^2} } 2\)

Square Root of Complex Number in Cartesian Form

\(\ds \)

\(=\)

\(\ds \dfrac {8 + \sqrt {64 + 80} } 2\)

\(\ds \)

\(=\)

\(\ds 4 + \sqrt {16 + 20}\)

\(\ds \)

\(=\)

\(\ds 4 + \sqrt {36}\)

\(\ds \)

\(=\)

\(\ds 10\)

\(\ds \leadsto \ \ \)

\(\ds x\)

\(=\)

\(\ds \pm \sqrt {10}\)

\(\ds \leadsto \ \ \)

\(\ds y\)

\(=\)

\(\ds \pm \dfrac {4 \sqrt 5} {2 \times \sqrt {10} }\)

\(\ds \)

\(=\)

\(\ds \pm \dfrac {2 \sqrt 5} {\sqrt 2 \times \sqrt 5}\)

\(\ds \)

\(=\)

\(\ds \pm \sqrt 2\)

As $2 x y = 4 \sqrt 5$ it follows that the two solutions are:

$\sqrt {10} + \sqrt 2 i$

$-\sqrt {10} - \sqrt 2 i$

$\blacksquare$

Sources

• 1981: Murray R. Spiegel: Theory and Problems of Complex Variables (SI ed.) ... (previous) ... (next): $1$: Complex Numbers: Supplementary Problems: Roots of Complex Numbers: $98 \ \text{(b)}$