18i complex number

Question

18i

Answer 1

Let √18i = a + bi, where a, b ∈ R.
Squaring on both sides, we get
18i = a² + b² i² + 2abi
0 + 18i = a² – b² + 2abi …..[∵ i² = -1]
Equating real and imaginary parts, we get
a² – b² = 0 and 2ab = 18
a² – b² = 0 and b = \(\frac{9}{a}\)
\(a^{2}-\left(\frac{9}{a}\right)^{2}=0\)
\(a^{2}-\frac{81}{a^{2}}=0\)
a⁴ – 81 = 0
(a² – 9) (a² + 9) = 0
a² = 9 or a² = -9
But a ∈ R
∴ a² ≠ -9
∴ a² = 9
∴ a = ± 3
When a = 3, b = \(\frac{9}{3}\) = 3
When a = -3, b = \(\frac{9}{-3}\) = -3
∴ √18i = ±(3 + 3i) = ±3(1 + i)