Question

Find the square roots of – 5 – 12i

Answer 1

Let `sqrt(- 5 - 12"i")` = a + ib

Squaring on both sides

– 5 – 12i = (a + ib)²

– 5 – 12i = a² – b² + 2iab

Equating real and imaginary parts

a² – b² = – 5

2ab = – 12

(a² + b²)² = (a² – b²)² + 4a²b²

= (– 5)² + (– 12)² = 169

∴ a² + b² = 13

Solving a² – b² = – 5 and a² + b² = 13

We get a² = 4, b² = 9

a = ± 2, b = ± 3

Since 2ab = – 12 < 0, a, b are of opposite signs.

∴ When a = ± 2, b = ± 3

Now `sqrt(- 5 - 12"i")` = ± (2 – 3i)

Aliter:

Square root of – 5 – 12i

Let a + ib = – 5 – 12i

a = – 5, b = – 12

|z| = `sqrt(5^2 + 12^2)`

= `sqrt(169)`

= 13

`sqrt("a" + "ib") = +-  [sqrt((sqrt("a"^2 + "b"^2) + "a")/2) + "i" "b"/"b"| sqrt((sqrt("a"^2 + "b"^2) - "a")/2)]`

= `+- [sqrt((13 - 5)/2) - "i" sqrt((13 + 5)/2)]`

= ± (2 – 3i).