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Find a+b When x²=3+4i and x³=2+11i | Complex Numbers

JEE Maths question with a full step-by-step solution.

Question
If x=a+ibx = a + ib is a complex number such that x2=3+4ix^2 = 3 + 4i and x3=2+11ix^3 = 2 + 11i, where i=−1i = \sqrt{-1}, then a+ba + b is
A11
B22
C33correct
D44
Solution
Step 1: Recover xx itself from the two given powers. Since x=x3x2x = \dfrac{x^3}{x^2}:
x=2+11i3+4ix = \frac{2 + 11i}{3 + 4i}
 Step 2: Rationalise by multiplying numerator and denominator by the conjugate 3−4i3 - 4i:
x=(2+11i)(3−4i)(3+4i)(3−4i)x = \frac{(2 + 11i)(3 - 4i)}{(3 + 4i)(3 - 4i)}
 Step 3: Expand the numerator:
(2+11i)(3−4i)=6−8i+33i−44i2=6+25i+44=50+25i(2 + 11i)(3 - 4i) = 6 - 8i + 33i - 44i^2 = 6 + 25i + 44 = 50 + 25i
 Compute the denominator:
(3+4i)(3−4i)=32+42=9+16=25(3 + 4i)(3 - 4i) = 3^2 + 4^2 = 9 + 16 = 25
 Step 4: Divide:
x=50+25i25=2+ix = \frac{50 + 25i}{25} = 2 + i
 Step 5: Identify a=2a = 2, b=1b = 1, so a+b=3a + b = 3. Correct answer: (3)
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