Find the square of 18i

1.	Find the square of 18i
Answer» 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 = \(\frac9a\) a² - (\(\frac9a\))² = 0 a² - \(\frac{81}{a^2}\) = 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 = 9/3 = 3 When a = -3, b = 9/-3 = -3 \(\therefore\) √18i = ±(3 + 3i) = ±3(1 + i)

Answer»

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 = \(\frac9a\)

a² - (\(\frac9a\))² = 0

a² - \(\frac{81}{a^2}\) = 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 = 9/3 = 3

When a = -3, b = 9/-3 = -3

\(\therefore\) √18i = ±(3 + 3i) = ±3(1 + i)

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