Find the Roots of Complex Numbers - Pre-Calculus

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Question

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

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Answer

Note that,

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

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Question

What is the length of

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

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Answer

We have

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

Hence,

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

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Question

Solve for $\small z^3=1$ (there may be more than one solution).

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Answer

Solving that equation is equivalent to solving the roots of the polynomial $\small f(z)=z^3-1$ .

Clearly, one of roots is 1.

Thus, we can factor the polynomial as $\small f(z)=(z-1)(z^2+z+1)$

so that we solve for the roots of $\small z^2+z+1$ .

Using the quadratic equation, we solve for roots, which are $\small z=\frac{-1+i\sqrt{3}}{2}, \frac{-1-i\sqrt{3}}{2}$ .

This means the solutions to $\small z^3=1$ are

$z=1, \frac{-1+i\sqrt{3}}{2}, \frac{-1-i\sqrt{3}}{2}$

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Question

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.

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Answer

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

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

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Question

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

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Answer

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

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Question

Compute

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Answer

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: $(3+3i)^5=(3\sqrt{2})^5\cdot e^{\frac{i5\pi}{4}=972\sqrt{2}(cos(\frac{5\pi}{4})+isin(\frac{5\pi}{4}))=972\sqrt{2}(\frac {-\sqrt{2}{2}}-\frac ({\sqrt{2}\cdot i}{2})=-972-972i$

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Question

Solve for all possible solutions to the quadratic expression:

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Answer

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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Question

Determine the length of

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Answer

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

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Question

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

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Answer

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

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Question

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

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Answer

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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Question

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

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Answer

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

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Question

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

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Answer

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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Question

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

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Answer

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.

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Question

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

Tap to reveal answer

Answer

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

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Question

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

Tap to reveal answer

Answer

Note that,

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

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Question

What is the length of

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

Tap to reveal answer

Answer

We have

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

Hence,

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

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Question

Solve for $\small z^3=1$ (there may be more than one solution).

Tap to reveal answer

Answer

Solving that equation is equivalent to solving the roots of the polynomial $\small f(z)=z^3-1$ .

Clearly, one of roots is 1.

Thus, we can factor the polynomial as $\small f(z)=(z-1)(z^2+z+1)$

so that we solve for the roots of $\small z^2+z+1$ .

Using the quadratic equation, we solve for roots, which are $\small z=\frac{-1+i\sqrt{3}}{2}, \frac{-1-i\sqrt{3}}{2}$ .

This means the solutions to $\small z^3=1$ are

$z=1, \frac{-1+i\sqrt{3}}{2}, \frac{-1-i\sqrt{3}}{2}$

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Question

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.

Tap to reveal answer

Answer

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

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

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Question

Compute

Tap to reveal answer

Answer

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: $(3+3i)^5=(3\sqrt{2})^5\cdot e^{\frac{i5\pi}{4}=972\sqrt{2}(cos(\frac{5\pi}{4})+isin(\frac{5\pi}{4}))=972\sqrt{2}(\frac {-\sqrt{2}{2}}-\frac ({\sqrt{2}\cdot i}{2})=-972-972i$

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Question

Solve for all possible solutions to the quadratic expression:

Tap to reveal answer

Answer

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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