One Problem at a Time

Verifying the Solutions of a Complex Equation

Problem 184: Verify that each of the two members z = 1 \pm i satisfies the equation z^2 - 2z + 2 = 0.

Solution: To verify the solution, we just need to plug in the solutions. That is,

If z = 1 + i, then

(1)   \begin{align*} z^2 - 2z + 2 & = 0 \\ (1+i)^2 - 2(1+i) + 2 & = 0 \\ (1+i)(1+i) - 2(1+i) + 2 & = 0 \\ 1 + i + i + i^2 -2 -2i + 2 & = 0 \\ 1 + i^2 & = 0 \\ 1 - 1 & = 0 \\ 0 & = 0. \end{align*}

Likewise, if z = 1 - i, then

(2)   \begin{align*} z^2 - 2z + 2 & = 0 \\ (1-i)^2 - 2(1-i) + 2 & = 0 \\ (1-i)(1-i) - 2(1-i) + 2 & = 0 \\ 1 - i - i + i^2 -2 +2i + 2 & = 0 \\ 1 + i^2 & = 0 \\ 1 - 1 & = 0 \\ 0 & = 0. \end{align*}

As desired, z =1 \pm i satisfies the equation z^2 - 2z + 2 = 0.

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