Let \(\sqrt{15-8i}=a+bi,\) where a, b b ∈ R.
Squaring on both sides, we get
15 – 8i = a2 + b2 i2 + 2abi
15 – 8i = (a2 – b2 ) + 2abi …..[∵ i = -1]
Equating real and imaginary parts, we get
a2 – b2 = 15 and 2ab = -8
a2 – b2 = 15 and b = \(\frac{-4}a\)
When a = 4, b = -4/4 = -1
When a = -4, b = -4/-4 = 1
\(\therefore\sqrt{15-8i}\) = ±(4 – i)