Precalculus : Powers and Roots of Complex Numbers

Study concepts, example questions & explanations for Precalculus

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Example Questions

Example Question #11 : Find The Roots Of Complex Numbers

Solve for \displaystyle z^{3}-z^{2}+z=0 (there may be more than one solution).

Possible Answers:

\displaystyle 0,\frac{1}{2},\pm \frac{\sqrt{3i}}{2}

\displaystyle 0,\frac{1}{2},\pm \sqrt{3i}

\displaystyle 0,1,\frac{1}{2}

\displaystyle 0

\displaystyle 0,\frac{1}{2},\pm \frac{\sqrt{3}}{2}

Correct answer:

\displaystyle 0,\frac{1}{2},\pm \frac{\sqrt{3i}}{2}

Explanation:

To solve for the roots, just set equal to zero and solve for z using the quadratic formula, which is \displaystyle \frac{-b\pm \sqrt{b^{2}4ac}}{2a}

\displaystyle z^{3}-z^{2}+z\rightarrow z\left ( z^{2}-z+1 \right ) and now setting both \displaystyle z and \displaystyle z^{2}-z+1 equal to zero we end up with the answers \displaystyle z=0 and \displaystyle z=\frac{1\pm 3i}{2} and so the correct answer is \displaystyle 0,\frac{1}{2},\pm \frac{2\sqrt{3i}}{2}.

Example Question #11 : Find The Roots Of Complex Numbers

Solve for all possible solutions to the quadratic expression: \displaystyle m^{2}-6m+12=0

Possible Answers:

\displaystyle 3\pm \sqrt{3}

\displaystyle 3\pm 3

\displaystyle 3\pm 3i

Explanation:

Solve for complex values of \displaystyle m using the quadratic formula: \displaystyle 6\pm\sqrt{36-(4)(1)(12)} = \frac{6\pm\sqrt{-12}}{2}=\frac{6\pm2\sqrt{3}i}{2}=3\pm\sqrt{3}i.

Example Question #11 : Find The Roots Of Complex Numbers

Solve for \displaystyle z^{3}-z^{2}+z=0 (there may be more than one solution).

Possible Answers:

\displaystyle 0,1,\frac{1}{2}

\displaystyle 0,\frac{1}{2},\pm \frac{\sqrt{3i}}{2}

\displaystyle 0,\frac{1}{2},\pm \frac{\sqrt{3}}{2}

\displaystyle 0,\frac{1}{2},\pm \sqrt{3i}

\displaystyle 0

Correct answer:

\displaystyle 0,\frac{1}{2},\pm \frac{\sqrt{3i}}{2}

Explanation:

To solve for the roots, just set equal to zero and solve for \displaystyle z using the quadratic formula (\displaystyle \frac{-b\pm \sqrt{b^{2}-4ac}}{2a}): \displaystyle z^{3}-z^{2}+z\rightarrow z(z^{2}-z+1) and now setting both \displaystyle z and \displaystyle z^{2}-z+1 equal to zero we end up with the answers \displaystyle z=0 and \displaystyle z=\frac{1\pm 3i}{2}.

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