Pre-Calculus - Find the Roots of Complex Numbers

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

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

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

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

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

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 and $z=\frac{1\pm 3i}{2}$ .

What is the length of

$({1-i})^{3}$ ?

$2\sqrt{2}$

$\sqrt{2}$

$2-\sqrt{2}$

$2+\sqrt{2}$

Explanation

We have

$|1-i|=\sqrt{2}$ .

Hence,

$|(1-i)^3|=|1-i|^{3}=(\sqrt2)^3=2\sqrt{2}$ .

Solve for all possible solutions to the quadratic expression:

$3\pm\sqrt{3}i$

$3\pm\sqrt{3}$

$3\pm3$

$3\pm3i$

Explanation

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

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

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

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

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

Solve for (there may be more than one solution).

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

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

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

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

Explanation

To solve for the roots, just set equal to zero and solve for z 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 equal to zero we end up with the answers and $z=\frac{1\pm 3i}{2}$

Recall that $z=rcis(\theta)$ is just shorthand for $z=r(cos(\theta)+isin(\theta))$ when dealing with complex numbers in polar form.

Express $\frac{1}{(\sqrt{3}-i)^4}$ in polar form.

$\frac{1}{16}cis\left(\frac{2\pi}{3}\right)$

$2cis\left(\frac{2\pi}{3}\right)$

$\frac{1}{16}cis\left(\frac{-2\pi}{3}\right)$

$2cis\left(\frac{\pi}{3}\right)$

$2cis\left(\frac{2\pi}{12}\right)$

Explanation

First we recognize that we are trying to solve $z^{-4}$ where $z=\sqrt{3}-i$ .

Then we want to convert into polar form using,

$arg(z)=\theta =arctan(\frac{b}{a})$ and $\left | z\right |=r=\sqrt{a^2+b^2}$ .

Then since De Moivre's theorem states,

$z^n=r^ncis(n\theta)$ if is an integer, we can say

$z^{-4}=2^{-4}cis(-4*\frac{-\pi}{6})=\frac{1}{16}cis\left(\frac{2\pi}{3}\right)$ .

Compute

Explanation

To solve this question, you must first derive a few values and convert the equation into exponential form: $re^{\pi*z}$ : $r=\sqrt{9+9}=3\sqrt{2}\rightarrow tan(\theta)=\frac{3}{3}=1\rightarrow Arg(z)=\frac{\pi}{4}$

Now plug back into the original equation and solve: $(3+3i)^5=(3\sqrt{2})^5\cdot e^{\frac{i5\pi}{4}=972\sqrt{2}(cos(\frac{5\pi}{4})+isin(\frac{5\pi}{4}))=972\sqrt{2}(\frac {-\sqrt{2}{2}}-\frac ({\sqrt{2}\cdot i}{2})=-972-972i$

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

Explanation

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

Evaluate ${}(1+i)^{2n}$ , where is a natural number and is the complex number ${}i^{2}=-1$ .

${}2^{n}i^{n}$

${}-2^{n}i^{n}$

${}2^{n}i^{n+1}$

${}2^{n+1}i^{n}$

${}-2^{n+1}i^{n}$

Explanation

Note that,

${}(1+i)^{2n}=((1+i)^2)^{n}}= (2i)^{n}= 2^{n}i^{n}$

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

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

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

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

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

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 and $z=\frac{1\pm 3i}{2}$ and so the correct answer is $0,\frac{1}{2},\pm \frac{2\sqrt{3i}}{2}$ .