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Precalculus : Polar Coordinates and Complex Numbers

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Example Questions

Example Question #2 : Find The Roots Of Complex Numbers

Determine the length of 2i^2-1

Possible Answers:

Correct answer:

Explanation:

i^2=-1 , so $2(-1)-1=-3\rightarrow |-3|=3$

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Example Question #2 : Find The Roots Of Complex Numbers

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

Possible Answers:

$3\pm3$

$3\pm\sqrt{3}$

$3\pm\sqrt{3}i$

$3\pm3i$

Correct answer:

$3\pm\sqrt{3}i$

Explanation:

Solve for complex values of m using the aforementioned quadratic formula:

$\frac{6\pm\sqrt{36-(4)(1)(12)}}{2}$

$=\frac{6\pm\sqrt{-12}}{2}$

$=\frac{6\pm2i\sqrt{3}}{2}$

$=3\pm \sqrt{3}i$

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Example Question #1 : Find The Roots Of Complex Numbers

Which of the following lists all possible solutions to the quadratic expression: $m^{2}-6m+12=0$

Possible Answers:

$3\pm 3i$

$3\pm \sqrt{3}$

$3\pm 3$

$3\pm \sqrt{3i}$

Correct answer:

$3\pm \sqrt{3i}$

Explanation:

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

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Example Question #10 : Find The Roots Of Complex Numbers

Determine the length of $2i^{2}-1$ .

Possible Answers:

Correct answer:

Explanation:

To begin, we must recall that $i^{2}=-1$ . Plug this in to get 2i^2-1=2(-1)-1=-2-1=-3 . Length must be a positive value, so we'll take the absolute value: $\left | -3 \right |=3$ . Therefore the length is 3.

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Example Question #11 : Find The Roots Of Complex Numbers

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

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

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

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Example Question #11 : Find The Roots Of Complex Numbers

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

Possible Answers:

$3\pm \sqrt{3}$

$3\pm \sqrt{3i}$

$3\pm 3$

$3\pm 3i$

Correct answer:

$3\pm \sqrt{3i}$

Explanation:

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

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Example Question #11 : Find The Roots Of Complex Numbers

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

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

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Example Question #1 : Express Complex Numbers In Rectangular Form

Convert the following to rectangular form:

2(cos(30)+ isin(30))

Possible Answers:

$\sqrt{3}+2i$

$\sqrt{3}-1$

$2\sqrt{3}+2i$

$2\sqrt{3}+1$

$\sqrt{3}+i$

Correct answer:

$\sqrt{3}+i$

Explanation:

Distribute the coefficient 2, and evaluate each term:

$\\2(cos(30)+ isin(30))\\ \\= 2cos(30)+i(2sin(30))\\ \\=\sqrt{3}+i$

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Example Question #1 : Express Complex Numbers In Rectangular Form

Convert the following to rectangular form: 3(cos(60)-isin(30))

Possible Answers:

$\frac{\sqrt3}{2}-\frac{i}{2}$

$\frac{3}{2}-\frac{3}{2}i$

$\frac{3\sqrt3}{2}-\frac{i}{2}$

$\frac{3}{2}-\frac{\sqrt3 i}{2}$

$\frac{3}{2}-\frac{i}{2}$

Correct answer:

$\frac{3}{2}-\frac{3}{2}i$

Explanation:

Distribute the coefficient and simplify:

$\\ 3(cos(60)-isin(30))\\ \\=3sin(60)-i(3sin(30)))\\ \\=\frac{3}{2}-\frac{3}{2}i$

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Example Question #1 : Polar Form Of Complex Numbers

Represent the polar equation:

$z = 3\left(cos(\frac{\pi}{6} )+i\cdot sin\left(\frac{\pi}{6}\right)\right)$

in rectangular form.

Possible Answers:

$z = cos\left(\frac{\pi}{6}\right) + sin\left(\frac{\pi}{6}\right)$

$z = 3cos\left(\frac{\pi}{6}\right) + 3sin\left(\frac{\pi}{6}\right)$

$z = \frac{3\sqrt{3}}{2} + \frac{3}{2}i$

$z=i+9\pi$

$z = \frac{2\sqrt{2}}{3} + \frac{2}{3}i$

Correct answer:

$z = \frac{3\sqrt{3}}{2} + \frac{3}{2}i$

Explanation:

Using the general form of a polar equation:

$z = r(cos(\theta) + i\cdot sin(\theta))$

we find that the value of is and the value of $\theta$ is $\frac{\pi}{6}$ .

The rectangular form of the equation appears as a + bi , and can be found by finding the trigonometric values of the cosine and sine equations.

$cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}$

$sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$

distributing the 3, we obtain the final answer of:

$\frac{3\sqrt{3}}{2} + \frac{3}{2}i$

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Precalculus : Polar Coordinates and Complex Numbers

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

Example Question #2 : Find The Roots Of Complex Numbers

Example Question #2 : Find The Roots Of Complex Numbers

Example Question #1 : Find The Roots Of Complex Numbers

Example Question #10 : Find The Roots Of Complex Numbers

Example Question #11 : Find The Roots Of Complex Numbers

Example Question #11 : Find The Roots Of Complex Numbers

Example Question #11 : Find The Roots Of Complex Numbers

Example Question #1 : Express Complex Numbers In Rectangular Form

Example Question #1 : Express Complex Numbers In Rectangular Form

Example Question #1 : Polar Form Of Complex Numbers

All Precalculus Resources

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