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(a+2i)(1+bi)=17-19 i

(a+2i)(1+bi)=17−19i (a+2 i)(1+b i)=17-19 i

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Full solution

Q. (a+2i)(1+bi)=17−19i (a+2 i)(1+b i)=17-19 i
  1. Expand left side: Expand the left side using the distributive property: (a+2i)(1+bi)=a+abi+2i+2bi2(a+2i)(1+bi) = a + abi + 2i + 2bi^2. Since i2=−1i^2 = -1, this simplifies to a+abi+2i−2ba + abi + 2i - 2b.
  2. Separate parts: Separate the real and imaginary parts: Real part: a−2ba - 2b, Imaginary part: abi+2iabi + 2i. Set these equal to the corresponding parts of the right side: a−2b=17a - 2b = 17 and abi+2=−19abi + 2 = -19.
  3. Solve real part: Solve the real part equation: a−2b=17a - 2b = 17.
  4. Solve imaginary part: Solve the imaginary part equation: abi+2=−19abi + 2 = -19. Since ii is not a real number, we can ignore the ii in abiabi for now and solve for bb: ab+2=−19ab + 2 = -19.

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