Imaginary Solution v3

Algebra Level 2

A tiny bit easier than before

x 2 ( x i + 1 ) ( x i − 3 ) 2 i x 2 = 4 i \frac{x^2\left(xi+1\right)\left(xi-3\right)}{2ix^2}=4i

What is the greatest value of ℑ ( x ) + ℜ ( x ) + 1 \Im(x)+\Re(x)+1 ?


The answer is 2.

Joshua Crawford by Joshua Crawford , 15, Australia

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1 solution

Chew-Seong Cheong
Mar 18, 2019

x 2 ( x i + 1 ) ( x i − 3 ) 2 i x 2 = 4 i For x ≠ 0 ( x i + 1 ) ( x i − 3 ) 2 i = 4 i Multiply both sides by 2 i ( x i + 1 ) ( x i − 3 ) = − 8 − x 2 − 2 x i − 3 = − 8 − x 2 − 2 x i + 5 = 0 x 2 + 2 x i − 5 = 0 ( x + 2 + i ) ( x − 2 + i ) = 0 \begin{aligned} \frac {x^2(xi+1)(xi-3)}{2ix^2} & = 4i & \small \color{#3D99F6} \text{For }x \ne 0 \\ \frac {(xi+1)(xi-3)}{2i} & = 4i & \small \color{#3D99F6} \text{Multiply both sides by } 2i \\ (xi+1)(xi-3) & = -8 \\ -x^2-2xi-3 & = -8 \\ -x^2-2xi+5 & = 0 \\ x^2+2xi-5 & = 0 \\ (x+2+i)(x-2+i) & = 0 \end{aligned}

⟹ { x = − 2 − i ⟹ ℑ ( x ) + ℜ ( x ) + 1 = − 2 − 1 + 1 = − 2 x = 2 − i ⟹ ℑ ( x ) + ℜ ( x ) + 1 = 2 − 1 + 1 = 2 \implies \begin{cases} x = -2-i & \implies \Im(x) + \Re(x) + 1 = -2-1+1 = - 2 \\ x = 2-i & \implies \Im(x) + \Re(x) + 1 = 2-1+1 = 2 \end{cases}

Therefore the maximum ℑ ( x ) + ℜ ( x ) + 1 \Im(x) + \Re(x) + 1 is 2 \boxed 2 .

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