Complex Numbers: Solve problems with complex numbers

Unit 1: Solve complex number problems

Dylan Busa

Unit 1 outcomes

By the end of this unit you will be able to:

• Solve for unknowns in equivalent complex numbers using simultaneous equation techniques.
• Solve quadratic equations that have complex roots.

What you should know

Before you start this unit, make sure you can:

Introduction

The quadratic function $\scriptsize f(x)={{x}^{2}}-2x+3$ (shown in Figure 1) does not intersect the x-axis and therefore has no real roots. But it does have roots. They are non-real or complex roots. What are the complex roots of the function?

To find the roots of $\scriptsize f(x)$, we know we need to solve for $\scriptsize x$ where $\scriptsize f(x)=0$. In other words, we need to solve the quadratic equation $\scriptsize {{x}^{2}}-2x+3=0$. Unfortunately, the quadratic expression on the left-hand side does not factorise easily. However, we can use the quadratic formula to solve for $\scriptsize x$.

Solve equations with complex roots

Let’s look at Example 1.1 to see how to solve for $\scriptsize x$ using the quadratic formula.

Example 1.1

Solve for $\scriptsize x$ in $\scriptsize {{x}^{2}}-2x+3=0$ using the quadratic formula.

Solution

$\scriptsize a=1,b=-2,c=3$

\scriptsize \begin{align*}x&=\displaystyle \frac{-b\pm\sqrt{b^2-4ac}}{2a}\\ &=\displaystyle \frac{-(-2)\pm\sqrt{(-2)^2-4(1)(3)}}{2(1)}\\ &=\displaystyle \frac{2\pm\sqrt{4-12}}{2}\\ &=\displaystyle \frac{2\pm\sqrt{-8}}{2}\\ &=\displaystyle \frac{2\pm\sqrt{8}i}{2}\\ &=\displaystyle \frac{2\pm2\sqrt{2}i}{2}\\ &=1\pm\sqrt{2}i \end{align*}

Therefore, $\scriptsize x=1+\sqrt{2}i$ or $\scriptsize x=1-\sqrt{2}i$.

We still get two roots (or solutions) as we expect from a quadratic equation. However, both roots are complex.

Note

The part of the quadratic formula under the square root sign ($\scriptsize {{b}^{2}}-4ac$) is called the discriminant. It is given the symbol $\scriptsize \Delta$ (Delta), which is the Greek letter D.

• If $\scriptsize \Delta \gt0$, then the roots of the quadratic equation are real and different.
• If $\scriptsize \Delta \lt0$, then the roots of the quadratic equation are complex and different.
• If $\scriptsize \Delta =0$, then the roots of the quadratic equation are real and the same.

Example 1.2

Solve for $\scriptsize m$ in $\scriptsize {{m}^{2}}-2m+5=0$.

Solution

The quadratic expression $\scriptsize m^2-2m+5$ does not factorise easily, so we will use the quadratic formula.
$\scriptsize a=1,b=-2,c=5$

\scriptsize \begin{align*}m&=\displaystyle \frac{{-b\pm \sqrt{{{{b}^{2}}-4ac}}}}{{2a}}\\&=\displaystyle \frac{{-(-2)\pm \sqrt{{{{{(-2)}}^{2}}-4(1)(5)}}}}{{2(1)}}\\&=\displaystyle \frac{{2\pm \sqrt{{4-20}}}}{2}\\&=\displaystyle \frac{{2\pm \sqrt{{-16}}}}{2}\\&=\displaystyle \frac{{2\pm 4i}}{2}\\&=1\pm 2i\end{align*}

Therefore, $\scriptsize m=1+2i$ or $\scriptsize m=1-2i$.

Example 1.3

Solve for $\scriptsize x$:
$\scriptsize \displaystyle \frac{3}{{x+3}}-\displaystyle \frac{2}{{x+2}}=1\quad \quad x\ne -3,x\ne -2$

Solution

We follow all the same steps as usual when solving equations, starting with multiplying through by the LCD.

\scriptsize \begin{align*}\displaystyle \frac{3}{{x+3}}-\displaystyle \frac{2}{{x+2}} & =1\quad \text{LCD: }(x+3)(x+2)\\\therefore 3(x+2)-2(x+3) & =(x+3)(x+2)\\\therefore 3x+6-2x-6 & ={{x}^{2}}+5x+6\\\therefore x & ={{x}^{2}}+5x+6\\\therefore {{x}^{2}}+4x+6 & =0\end{align*}

The quadratic expression does not factorise easily, so we will use the quadratic formula.
$\scriptsize a=1,b=4,c=6$

\scriptsize \begin{align*}x&=\displaystyle \frac{{-b\pm \sqrt{{{{b}^{2}}-4ac}}}}{{2a}}\\&=\displaystyle \frac{{-(4)\pm \sqrt{{{{{(4)}}^{2}}-4(1)(6)}}}}{{2(1)}}\\&=\displaystyle \frac{{-4\pm \sqrt{{16-24}}}}{2}\\&=\displaystyle \frac{{-4\pm \sqrt{{-8}}}}{2}\\&=\displaystyle \frac{{-4\pm 2\sqrt{2}i}}{2}\\&=-2\pm \sqrt{2}i\end{align*}

Therefore, $\scriptsize x=-2+\sqrt{2}i$ or $\scriptsize x=-2-\sqrt{2}i$.

Exercise 1.1

Solve for $\scriptsize x$, expressing any complex roots in standard form:

1. $\scriptsize {{x}^{2}}+x+1=0$
2. $\scriptsize 5{{x}^{2}}-8x=-6$
3. $\scriptsize 2{{x}^{2}}-12x+19=0$
4. $\scriptsize \displaystyle \frac{x}{{(x-2)}}+\displaystyle \frac{3}{{(x-3)}}=3$
5. $\scriptsize {{x}^{2}}+{{(x+1)}^{2}}+{{(x+2)}^{2}}=-1$
6. $\scriptsize {{x}^{4}}=-8{{x}^{2}}$

The full solutions are at the end of the unit.

Solve complex simultaneous equations

We know that complex numbers have a real and imaginary part. If we have an equation such as $\scriptsize 2x+3yi=1+i$, we can easily find the values for $\scriptsize x$ and $\scriptsize y$ that will make the equation true. We simply need to equate the real and imaginary parts.
\scriptsize \begin{align*}2x & =1\\\therefore x & =\displaystyle \frac{1}{2}\end{align*}
\scriptsize \begin{align*}3y & =1\\\therefore y & =\displaystyle \frac{1}{3}\end{align*}

Sometimes we find $\scriptsize x$ and $\scriptsize y$ in both the real and imaginary parts. For example, if we wanted to find the values of $\scriptsize x$ and $\scriptsize y$ in $\scriptsize (x-y)-(x+y)i=2+3i$, we could set up a system of two simultaneous equations and then solve for the unknowns.

Have a look at the next example to see how to answer this question.

Example 1.4

Solve for $\scriptsize x$ and $\scriptsize y$ in the following:
$\scriptsize (x-y)-(x+y)i=2+3i$

Solution

On the left-hand side (LHS), the real part is $\scriptsize (x-y)$ and the imaginary part is $\scriptsize -(x+y)i$. Therefore $\scriptsize a=(x-y)$ and $\scriptsize b=-(x+y)$.

On the right-hand side (RHS) we have another complex number where $\scriptsize a=2$ and $\scriptsize b=3$.

But because the complex number on the LHS is equal to the complex number on the RHS, we know that real and imaginary parts must be equal, in other words, that $\scriptsize (x-y)=2$ and $\scriptsize -(x+y)=3$. We have a system of simultaneous equations that we can use to solve for $\scriptsize x$ and $\scriptsize y$.

$\scriptsize (x-y)=2\quad \quad (1)$
$\scriptsize -(x+y)=3\quad \quad (2)$

From $\scriptsize (1)$:
\scriptsize \begin{align*}(x-y) & =2\\\therefore x & =2+y\quad \quad (3)\end{align*}
Substitute $\scriptsize (3)$ into $\scriptsize (2)$:
\scriptsize \begin{align*}-\left( {(2+y)+y} \right) & =3\\\therefore -(2+2y) & =3\\\therefore -2-2y & =3\\\therefore 2y & =-5\\\therefore y & =-\displaystyle \frac{5}{2}\end{align*}
Substitute $\scriptsize y=-\displaystyle \frac{5}{2}$ into $\scriptsize (3)$:
\scriptsize \begin{align*}x & =2-\displaystyle \frac{5}{2}\\\therefore x & =-\displaystyle \frac{1}{2}\end{align*}

$\scriptsize x =-\displaystyle \frac{1}{2}$ and $\scriptsize y=-\displaystyle \frac{5}{2}$

Example 1.5

Solve for $\scriptsize x$ and $\scriptsize y$ in $\scriptsize \displaystyle \frac{{2i-4}}{{3-2i}}=\displaystyle \frac{{i-x}}{{y-1}}$.

Solution

We need to simplify first in order to set up a system of simultaneous equations.
\scriptsize \begin{align*}\displaystyle \frac{{2i-4}}{{3-2i}} & =\displaystyle \frac{{i-x}}{{y-1}}&& y\ne 1 \text{ LCD: }(3-2i)(y-1)\\\therefore (2i-4)(y-1) & =(i-x)(3-2i)\\\therefore 2iy-2i-4y+4 & =3i-2{{i}^{2}}-3x-2xi&& \text{Collect the real and imaginary parts on each side}\\\therefore (4-4y)+(2y-2)i & =(2-3x)+(3-2x)i\end{align*}

Now we can set up a system of simultaneous equations by equating the real and imaginary parts of the complex numbers.
\scriptsize \begin{align*}(4-4y) &=(2-3x)\\\therefore 4-4y & =2-3x\\\therefore 4y-4 & =3x-2\\\therefore 4y & =3x+2\\\therefore y & =\displaystyle \frac{3}{4}x+\displaystyle \frac{1}{2}\quad \quad (1)\end{align*}

\scriptsize \begin{align*}(2y-2) & =(3-2x)\\\therefore 2y-2 & =3-2x\\\therefore 2y & =5-2x\\\therefore y & =\displaystyle \frac{5}{2}-x\quad \quad (2)\end{align*}

Substitute $\scriptsize (1)$ into $\scriptsize (2)$:
\scriptsize \begin{align*}\displaystyle \frac{3}{4}x+\displaystyle \frac{1}{2} & =\displaystyle \frac{5}{2}-x\\\therefore 3x+2 & =10-4x\\\therefore 7x & =8\\\therefore x & =\displaystyle \frac{8}{7}\end{align*}

Substitute $\scriptsize x =\displaystyle \frac{8}{7}$ into $\scriptsize (2)$:
\scriptsize \begin{align*}y&=\displaystyle \frac{5}{2}-\displaystyle \frac{8}{7}\\&=\displaystyle \frac{{35-16}}{{14}}\\&=\displaystyle \frac{{19}}{{14}}\end{align*}

$\scriptsize x =\displaystyle \frac{8}{7}$ and $\scriptsize y=\displaystyle \frac{{19}}{{14}}$

Exercise 1.2

Solve for $\scriptsize x$ and $\scriptsize y$ in each of the following:

1. $\scriptsize 3x+4yi=\sqrt{3}-\sqrt{3}i$
2. $\scriptsize (2x-y)+(x-2y)i=4-2i$
3. $\scriptsize \displaystyle \frac{{-2+i}}{{2+i}}=\displaystyle \frac{{2i-3x}}{{y-1}}$

The full solutions are at the end of the unit.

Summary

In this unit you have learnt the following:

• How to solve equations where some of the roots are complex.
• How to solve for unknowns in equivalent complex numbers using the technique of simultaneous equations.

Unit 1: Assessment

Suggested time to complete: 15 minutes

1. Solve for $\scriptsize x$, leaving any complex roots in standard form:
1. $\scriptsize 2{{x}^{2}}+4x=-11$
2. $\scriptsize -{{x}^{2}}+x-23=0$
3. $\scriptsize -3{{x}^{2}}+2x=14$
4. $\scriptsize \displaystyle \frac{1}{2}{{x}^{2}}+4x=-12$
5. $\scriptsize 4{{x}^{2}}=-{{x}^{4}}$
6. $\scriptsize \displaystyle \frac{4}{{(x+1)}}-\displaystyle \frac{3}{x}=7$
2. Solve for $\scriptsize x$ and $\scriptsize y$ in the following equations:
1. $\scriptsize (x+y)-(x-y)i=3+2i$
2. $\scriptsize \displaystyle \frac{{3i-2}}{{3+2i}}=\displaystyle \frac{{i+x}}{{y-1}}$
3. $\scriptsize x+yi=\displaystyle \frac{{{{{(2-3i)}}^{2}}}}{{1+i}}$
3. Show that $\scriptsize (5+4i)$ and $\scriptsize (3+2i)$ are factors of $\scriptsize (7+22i)$. Hence, or otherwise, determine the prime factors of $\scriptsize ({{7}^{2}}+{{22}^{2}})$.

The full solutions are at the end of the unit.

Unit 1: Solutions

Exercise 1.1

1. $\scriptsize {{x}^{2}}+x+1=0$
$\scriptsize a=1,b=1,c=1$
\scriptsize \begin{align*}x&=\displaystyle \frac{{-b\pm \sqrt{{{{b}^{2}}-4ac}}}}{{2a}}\\&=\displaystyle \frac{{-(1)\pm \sqrt{{{{{(1)}}^{2}}-4(1)(1)}}}}{{2(1)}}\\&=\displaystyle \frac{{-1\pm \sqrt{{1-4}}}}{2}\\&=\displaystyle \frac{{-1\pm \sqrt{{-3}}}}{2}\\&=-\displaystyle \frac{1}{2}\pm \displaystyle \frac{{\sqrt{3}}}{2}i\end{align*}
$\scriptsize x=-\displaystyle \frac{1}{2}+\displaystyle \frac{{\sqrt{3}}}{2}i$ or $\scriptsize x=-\displaystyle \frac{1}{2}-\displaystyle \frac{{\sqrt{3}}}{2}i$
2. .
\scriptsize \begin{align*}5{{x}^{2}}-8x & =-6\\\therefore 5{{x}^{2}}-8x+6 & =0\end{align*}
$\scriptsize a=5,b=-8,c=6$
\scriptsize \begin{align*}x&=\displaystyle \frac{{-b\pm \sqrt{{{{b}^{2}}-4ac}}}}{{2a}}\\&=\displaystyle \frac{{-(-8)\pm \sqrt{{{{{(-8)}}^{2}}-4(5)(6)}}}}{{2(5)}}\\&=\displaystyle \frac{{8\pm \sqrt{{64-120}}}}{{10}}\\&=\displaystyle \frac{{8\pm \sqrt{{-56}}}}{{10}}\\&=\displaystyle \frac{{8\pm \sqrt{{56}}i}}{{10}}\\&=\displaystyle \frac{{8\pm 2\sqrt{{14}}i}}{{10}}\\&=\displaystyle \frac{4}{5}\pm \displaystyle \frac{{\sqrt{{14}}}}{5}i\end{align*}
$\scriptsize x=\displaystyle \frac{4}{5}+\displaystyle \frac{{\sqrt{{14}}}}{5}i$ or $\scriptsize x=\displaystyle \frac{4}{5}-\displaystyle \frac{{\sqrt{{14}}}}{5}i$
3. $\scriptsize 2{{x}^{2}}-12x+19=0$
$\scriptsize a=2,b=-12,c=19$
\scriptsize \begin{align*}x&=\displaystyle \frac{{-b\pm \sqrt{{{{b}^{2}}-4ac}}}}{{2a}}\\&=\displaystyle \frac{{-(-12)\pm \sqrt{{{{{(-12)}}^{2}}-4(2)(19)}}}}{{2(2)}}\\&=\displaystyle \frac{{12\pm \sqrt{{144-152}}}}{4}\\&=\displaystyle \frac{{12\pm \sqrt{{-8}}}}{4}\\&=\displaystyle \frac{{12\pm 2\sqrt{2}i}}{4}\\&=3\pm \displaystyle \frac{{\sqrt{2}}}{2}i\end{align*}
$\scriptsize x=3+\displaystyle \frac{{\sqrt{2}}}{2}i$ or $\scriptsize x=3-\displaystyle \frac{{\sqrt{2}}}{2}i$
4. .
\scriptsize \begin{align*}\displaystyle \frac{x}{{(x-2)}}+\displaystyle \frac{3}{{(x-3)}} & =3\quad \quad x\ne 2,x\ne 3\quad \text{LCD: }(x-2)(x-3)\\\therefore x(x-3)+3(x-2) & =3(x-3)(x-2)\\\therefore {{x}^{2}}-3x+3x-6 & =3{{x}^{2}}-15x+18\\\therefore 2{{x}^{2}}-15x+24 & =0\end{align*}
$\scriptsize a=2,b=-15,c=24$
\scriptsize \begin{align*}x&=\displaystyle \frac{{-b\pm \sqrt{{{{b}^{2}}-4ac}}}}{{2a}}\\&=\displaystyle \frac{{-(-15)\pm \sqrt{{{{{(-15)}}^{2}}-4(2)(24)}}}}{{2(2)}}\\&=\displaystyle \frac{{15\pm \sqrt{{225-192}}}}{4}\\&=\displaystyle \frac{{15\pm \sqrt{{33}}}}{4}\end{align*}
$\scriptsize x=\displaystyle \frac{{15+\sqrt{{33}}}}{4}$ or $\scriptsize x=\displaystyle \frac{{15-\sqrt{{33}}}}{4}$
5. .
\scriptsize \begin{align*}x^2+(x+1)^2+(x+2)^2&=-1\\ \therefore x^2+x^2+2x+1+x^2+4x+4&=-1\\ \therefore 3x^2+6x+6=0\end{align*}
$\scriptsize a=3,b=6,c=6$
\scriptsize \begin{align*}x&=\displaystyle \frac{{-b\pm \sqrt{{{{b}^{2}}-4ac}}}}{{2a}}\\&=\displaystyle \frac{{-(6)\pm \sqrt{{{{{(6)}}^{2}}-4(3)(6)}}}}{{2(3)}}\\&=\displaystyle \frac{{-6\pm \sqrt{{36-72}}}}{6}\\&=\displaystyle \frac{{-6\pm \sqrt{{-36}}}}{6}\\&=\displaystyle \frac{{-6\pm 6i}}{6}\\&=-1\pm i\end{align*}
$\scriptsize x=-1+i$ or $\scriptsize x=-1-i$
6. .
\scriptsize \begin{align*}x^4&=-8x^2\\ \therefore x^4+8x^2&=0\\ \therefore x^2(x^2+8)&=0\\ \therefore x^2=0&\text{ or }x^2=-8\\ \therefore x=0&\text{ or }x=\pm\sqrt{8}i\\ \therefore x=0&\text{ or }x=\pm2\sqrt{2}i\end{align*}/li>

Back to Exercise 1.1

Exercise 1.2

1. $\scriptsize 3x+4yi=\sqrt{3}-\sqrt{3}i$
\scriptsize \begin{align*}3x & =\sqrt{3}\\\therefore x & =\displaystyle \frac{{\sqrt{3}}}{3}\end{align*}
\scriptsize \begin{align*}4y & =-\sqrt{3}\\\therefore y & =-\displaystyle \frac{{\sqrt{3}}}{4}\end{align*}
2. $\scriptsize (2x-y)+(x-2y)i=4-2i$
$\scriptsize 2x-y=4\quad \quad (1)$
$\scriptsize x-2y=-2\quad \quad (2)$
From $\scriptsize (2)$:
$\scriptsize x=2y-2\quad \quad (3)$
Substitute $\scriptsize (3)$ into $\scriptsize (1)$:
\scriptsize \begin{align*}2(2y-2)-y & =4\\\therefore 4y-4-y & =4\\\therefore 3y & =8\\\therefore y & =\displaystyle \frac{8}{3}\end{align*}
Substitute $\scriptsize y=\displaystyle \frac{8}{3}$ into $\scriptsize (1)$:
\scriptsize \begin{align*}2x-\displaystyle \frac{8}{3} & =4\\\therefore 2x & =4+\displaystyle \frac{8}{3}\\ & =\displaystyle \frac{{20}}{3}\end{align*}
$\scriptsize x=\displaystyle \frac{{20}}{3}$ and $\scriptsize y=\displaystyle \frac{8}{3}$
3. .
\scriptsize \begin{align*}\displaystyle \frac{-2+i}{2+i}&=\displaystyle \frac{2i-3x}{y-1}&&y\ne1\text{ LCD: }(2+i)(y-1)\\ \therefore (-2+i)(y-1)&=(2i-3x)(2+i)\\ \therefore -2y+2+yi-i&=4i+2i^2-6x-3xi&&\text{Collect real amd imaginary terms on both sides}\\ \therefore (2-2y)+(y-1)i&=(-2-6x)+(4-3x)i\end{align*}The real parts on both sides are equal and the imaginary parts on both sides are equal.

\scriptsize \begin{align*}(2-2y) & =(-2-6x)\\\therefore 2-2y & =-2-6x\\\therefore 2y & =6x+4\\\therefore y & =3x+2\quad \quad (1)\end{align*}

\scriptsize \begin{align*}(y-1)&=(4-3x)\\\therefore y-1&=4-3x\\\therefore y & =5-3x\quad \quad (2)\end{align*}

Substitute $\scriptsize (1)$ into $\scriptsize (2)$:
\scriptsize \begin{align*}3x+2 & =5-3x\\\therefore 6x & =3\\\therefore x & =\displaystyle \frac{1}{2}\end{align*}

Substitute $\scriptsize x=\displaystyle \frac{1}{2}$ into $\scriptsize (1)$:
\scriptsize \begin{align*}y & =3\left( {\displaystyle \frac{1}{2}} \right)+2\\ & =\displaystyle \frac{3}{2}+2\\ & =\displaystyle \frac{7}{2}\end{align*}
$\scriptsize x=\displaystyle \frac{1}{2}$ and $\scriptsize y =\displaystyle \frac{7}{2}$

Back to Exercise 1.2

Unit 1: Assessment

1. .
1. .
\scriptsize \begin{align*}2x^2+4x&=-11\\\therefore2x^2+4x+11&=0\end{align*}
$\scriptsize a=2,b=4,c=11$
\scriptsize \begin{align*}x&=\displaystyle \frac{{-b\pm \sqrt{{{{b}^{2}}-4ac}}}}{{2a}}\\&=\displaystyle \frac{{-(4)\pm \sqrt{{{{{(4)}}^{2}}-4(2)(11)}}}}{{2(2)}}\\&=\displaystyle \frac{{-4\pm \sqrt{{16-88}}}}{4}\\&=\displaystyle \frac{{-4\pm \sqrt{{-72}}}}{4}\\&=\displaystyle \frac{{-4\pm \sqrt{{72}}i}}{4}\\&=\displaystyle \frac{{-4\pm 6\sqrt{2}i}}{4}\\&=-1\pm \displaystyle \frac{{3\sqrt{2}}}{2}i\end{align*}
$\scriptsize x=-1+\displaystyle \frac{{3\sqrt{2}}}{2}i$ or $\scriptsize x=-1-\displaystyle \frac{{3\sqrt{2}}}{2}i$
2. .
\scriptsize \begin{align*}-x^2+x-23&=0\\\therefore x^2-x+23&=0\end{align*}
$\scriptsize a=1,b=-1,c=23$
\scriptsize \begin{align*}x&=\displaystyle \frac{{-b\pm \sqrt{{{{b}^{2}}-4ac}}}}{{2a}}\\&=\displaystyle \frac{{-(-1)\pm \sqrt{{{{{(-1)}}^{2}}-4(1)(23)}}}}{{2(1)}}\\&=\displaystyle \frac{{1\pm \sqrt{{1-92}}}}{2}\\&=\displaystyle \frac{{1\pm \sqrt{{-91}}}}{2}\\&=\displaystyle \frac{{1\pm \sqrt{{91}}i}}{2}\\&=\displaystyle \frac{1}{2}\pm \displaystyle \frac{{\sqrt{{91}}}}{2}i\end{align*}
$\scriptsize x=\displaystyle \frac{1}{2}+\displaystyle \frac{{\sqrt{{91}}}}{2}i$ or $\scriptsize x=\displaystyle \frac{1}{2}-\displaystyle \frac{{\sqrt{{91}}}}{2}i$
3. .
\scriptsize \begin{align*}-3x^2+2x&=14\\\therefore 3x^2-2x+14&=0\end{align*}
$\scriptsize a=3,b=-2,c=14$
\scriptsize \begin{align*}x&=\displaystyle \frac{{-b\pm \sqrt{{{{b}^{2}}-4ac}}}}{{2a}}\\&=\displaystyle \frac{{-(-2)\pm \sqrt{{{{{(-2)}}^{2}}-4(3)(14)}}}}{{2(3)}}\\&=\displaystyle \frac{{2\pm \sqrt{{4-168}}}}{6}\\&=\displaystyle \frac{{2\pm \sqrt{{-164}}}}{6}\\&=\displaystyle \frac{{2\pm \sqrt{{164}}i}}{6}\\&=\displaystyle \frac{{2\pm 2\sqrt{{41}}i}}{6}\\&=\displaystyle \frac{1}{3}\pm \displaystyle \frac{{\sqrt{{41}}}}{3}i\end{align*}
$\scriptsize x=\displaystyle \frac{1}{3}+\displaystyle \frac{{\sqrt{{41}}}}{3}i$ or $\scriptsize x=\displaystyle \frac{1}{3}-\displaystyle \frac{{\sqrt{{41}}}}{3}i$
4. .
\scriptsize \begin{align*}\displaystyle \frac{1}{2}{{x}^{2}}+4x & =-12\\\therefore {{x}^{2}}+8x+24 & =0\end{align*}
$\scriptsize a=1,b=8,c=24$
\scriptsize \begin{align*}x&=\displaystyle \frac{{-b\pm \sqrt{{{{b}^{2}}-4ac}}}}{{2a}}\\&=\displaystyle \frac{{-(8)\pm \sqrt{{{{{(8)}}^{2}}-4(1)(24)}}}}{{2(1)}}\\&=\displaystyle \frac{{-8\pm \sqrt{{64-96}}}}{2}\\&=\displaystyle \frac{{-8\pm \sqrt{{-32}}}}{2}\\&=\displaystyle \frac{{-8\pm \sqrt{{32}}i}}{2}\\&=\displaystyle \frac{{-8\pm 4\sqrt{2}i}}{2}\\&=-4\pm 2\sqrt{2}i\end{align*}
$\scriptsize x=-4+2\sqrt{2}i$ or $\scriptsize x=-4-2\sqrt{2}i$
5. .
\scriptsize \begin{align*}4x^2&=-x^4\\ \therefore x^4+4x^2&=0\\ \therefore x^2(x^2+4)&=0\\ \therefore x^2=0&\text{ or }x^2=-4\\ \therefore x=0&\text{ or }x=\pm\sqrt{-4}\\ \therefore x=0&\text{ or }x=\pm2i\end{align*}
6. .
\scriptsize \begin{align*}\displaystyle \frac{4}{(x+1)}-\displaystyle \frac{3}{x}&=7&&x\ne-1,x\ne0\text{ LCD: }x(x+1)\\ \therefore 4x-3(x+1)&=7x(x+1)\\ \therefore 4x-3x-3&=7x^2+7x\\ \therefore 7x^2+6x+3&=0\end{align*}
$\scriptsize a=7,b=6,c=3$
\scriptsize \begin{align*}x&=\displaystyle \frac{{-b\pm \sqrt{{{{b}^{2}}-4ac}}}}{{2a}}\\&=\displaystyle \frac{{-(6)\pm \sqrt{{{{{(6)}}^{2}}-4(7)(3)}}}}{{2(7)}}\\&=\displaystyle \frac{{-6\pm \sqrt{{36-84}}}}{{14}}\\&=\displaystyle \frac{{-6\pm \sqrt{{-48}}}}{{14}}\\&=\displaystyle \frac{{-6\pm \sqrt{{48}}i}}{{14}}\\&=\displaystyle \frac{{-6\pm 4\sqrt{3}i}}{{14}}\\&=-\displaystyle \frac{3}{7}\pm \displaystyle \frac{{2\sqrt{3}}}{7}i\end{align*}
$\scriptsize x=-\displaystyle \frac{3}{7}+\displaystyle \frac{{2\sqrt{3}}}{7}i$ or $\scriptsize x=-\displaystyle \frac{3}{7}-\displaystyle \frac{{2\sqrt{3}}}{7}i$
2. .
1. $\scriptsize (x+y)-(x-y)i=3+2i$\scriptsize \begin{align*}x+y & =3\\\therefore x & =3-y\quad \quad (1)\end{align*}

\scriptsize \begin{align*}-(x-y) & =2\\\therefore -x+y & =2\\\therefore x & =y-2\quad \quad (2)\end{align*}

Substitute $\scriptsize (1)$ into $\scriptsize (2)$:
\scriptsize \begin{align*}3-y & =y-2\\\therefore 2y & =5\\\therefore y & =\displaystyle \frac{5}{2}\end{align*}

Substitute $\scriptsize y =\displaystyle \frac{5}{2}$ into $\scriptsize (1)$:
\scriptsize \begin{align*}x & =3-\displaystyle \frac{5}{2}\\\therefore x & =\displaystyle \frac{1}{2}\end{align*}
$\scriptsize x=\displaystyle \frac{1}{2}$ and $\scriptsize y=\displaystyle \frac{5}{2}$

2. .
\scriptsize \begin{align*}\displaystyle \frac{3i-2}{3+2i}&=\displaystyle \frac{i+x}{y-1}&&y\ne1\text{ LCD: }(3+2i)(y-1)\\ \therefore (3i-2)(y-1)&=(i+x)(3+2i)\\ \therefore 3yi-3i-2y+2&=3i+2i^2+3x+2xi\\ \therefore (2-2y)+(3y-3)i&=(3x-2)+(3+2x)i\end{align*}\scriptsize \begin{align*}(2-2y) & =(3x-2)\\\therefore 2-2y & =3x-2\\\therefore 2y & =-3x+4\\\therefore y & =-\displaystyle \frac{3}{2}x+2\quad \quad (1)\end{align*}

\scriptsize \begin{align*}(3y-3) & =(3+2x)\\\therefore 3y-3 & =3+2x\\\therefore 3y & =2x+6\\\therefore y & =\displaystyle \frac{2}{3}x+2\quad (2)\end{align*}

Substitute $\scriptsize (1)$ into $\scriptsize (2)$:
\scriptsize \begin{align*}-\displaystyle \frac{{3x}}{2}+2 & =\displaystyle \frac{{2x}}{3}+2\\\therefore -9x+12 & =4x+12\\\therefore 13x & =0\\\therefore x & =0\end{align*}

Substitute $\scriptsize x=0$ into $\scriptsize (1)$:
$\scriptsize y=2$
$\scriptsize x=0$ and $\scriptsize y=2$

3. .
\scriptsize \begin{align*}x+yi & =\displaystyle \frac{{{{{(2-3i)}}^{2}}}}{{1+i}}\\\therefore (x+yi)(1+i) & =(2-3i)(2-3i)\\\therefore x+xi+yi+y{{i}^{2}} & =4-6i-6i+9{{i}^{2}}\\\therefore (x-y)+(x+y)i & =-5-12i\end{align*}\scriptsize \begin{align*}x-y & =-5\\\therefore x & =y-5\quad \quad (1)\end{align*}\scriptsize \begin{align*}x+y & =-12\\\therefore x & =-y-12\quad \quad (2)\end{align*}

Substitute $\scriptsize (1)$ into $\scriptsize (2)$:
\scriptsize \begin{align*}y-5 & =-y-12\\\therefore 2y & =-7\\\therefore y & =-\displaystyle \frac{7}{2}\end{align*}

Substitute $\scriptsize y =-\displaystyle \frac{7}{2}$ into $\scriptsize (1)$:
\scriptsize \begin{align*}x & =-\displaystyle \frac{7}{2}-7\\=\displaystyle \frac{{-7-14}}{2}\\=-\displaystyle \frac{{21}}{2}\end{align*}

$\scriptsize x=-\displaystyle \frac{{21}}{2}$ and $\scriptsize y=-\displaystyle \frac{7}{2}$

3. .
\scriptsize \begin{align*}(5+4i)\times (3+2i)&=15+10i+12i+8{{i}^{2}}\\&=15+22i-8\\&=7+22i\end{align*}Therefore $\scriptsize (5+4i)$ and $\scriptsize (3+2i)$ are factors of $\scriptsize (7+22i)$.

\scriptsize \begin{align*}({{7}^{2}}+{{22}^{2}})&={{7}^{2}}-{{22}^{2}}{{i}^{2}}\\&={{7}^{2}}-{{(22i)}^{2}}\\&=(7+22i)(7-22i)\end{align*}

Now:
\scriptsize \begin{align*}(5-4i)\times (3-2i)&=15-10i-12i+8{{i}^{2}}\\&=15-22i-8\\&=7-22i\end{align*}
And $\scriptsize (7+22i)=(5+4i)(3+2i)$ (from above)

Therefore:
\scriptsize \begin{align*}({{7}^{2}}+{{22}^{2}})&=(7+22i)(7-22i)\\&=(5+4i)(3+2i)(5-4i)(3-2i)\\&=\left( {(5+4i)(5-4i)} \right)\left( {(3+2i)(3-2i)} \right)\\&=(25-16{{i}^{2}})(9-4{{i}^{2}})\\&=(25+16)(9+4)\\&=(41)(13)\end{align*}

Therefore, the prime factors of $\scriptsize ({{7}^{2}}+{{22}^{2}})$ are $\scriptsize 41$ and $\scriptsize 13$.

Back to Unit 1: Assessment