Complex Conjugates

Given a complex number $z = a + bi \,(a, b \in \mathbb{R})$, the complex conjugate of $z,$ denoted $\overline{z},$ is the complex number $\overline{z} = a - bi$. The complex conjugate has the same real component $a$, but has opposite sign for the imaginary component $b$.

Properties of complex conjugates:

1. $\hspace{1mm} \overline { z+w }=\overline {z} +\overline {w}$
2. $\hspace{1mm} \overline { z-w }=\overline { z } -\overline { w }$
3. $\hspace{1mm} \overline { z\cdot w } =\overline { z } \cdot \overline { w }$
4. $\hspace{1mm} \overline{\left(\frac zw\right)}=\frac { \overline { z } }{ \overline { w } }$
5. $\hspace{1mm} \overline { \overline { z } } = z,$ i.e. the conjugate of the conjugate of $z$ is $z.$
6. $\hspace{1mm} \overline { z } = z$, if and only if $z$ is real.
7. $\hspace{1mm} { |z| }^{ 2 }=z\cdot \overline { z } =\overline { z } \cdot z$, where $|z|$ stands for the modulus of $z.$
8. $\hspace{1mm} \overline { { z }^{ n }}= \left(\overline{ z }\right)^{ n }$

Operations on $z$ and $\overline {z}:$

• a) $z+\overline{z}=(a+ib)+(a-ib)=2a$ $(\in \mathbb{R})$
• b) $z-\overline { z } =(a+ib)-(a-ib)=2bi$ (pure imaginary number)
• c) $z\cdot \overline { z } =(a+ib)(a-ib)={ a }^{ 2 }-abi+abi-{ b }^{ 2 }{ i }^{ 2 }={ a }^{ 2 }+{ b }^{ 2 }$ $\Big({ i }^{ 2 }=\big({ \sqrt { -1 } }\big)^{ 2 }=-1\Big)$

Based on these operations, we can add some more properties of conjugate:

$\hspace{1mm}$ 9. $\hspace{1mm} z+\overline{z}=2\text{Re}(z)$, twice the real element of $z.$
$\hspace{1mm}$ 10. $\hspace{1mm} z-\overline { z } =2\text{Im}(z)$, twice the imaginary element of $z.$

Why conjugate?

The need of conjugation comes from the fact that ${ i }^{ 2 }=-1$. This means that the equation has two roots, namely $i$ and $-i$.

What this tells us is that from the standpoint of real numbers, both are indistinguishable. For example, for a polynomial $f(x)$ with real coefficient, $f(z=a+bi)=0$ could be a solution if and only if its conjugate is also a solution $f(\overline z=a-bi)=0$. This means they are basically the same in the real numbers frame.

One importance of conjugation comes from the fact the product of a complex number with its conjugate, is a real number!! (See the operation c) above.) This can come in handy when simplifying complex expressions. It is like rationalizing a rational expression. Let's look at an example to see what we mean.

Perform the necessary operations to put $\frac{4+3i}{5+2i}$ to $a+bi \,(a,b \in \mathbb{R})$ form.

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Multiply both the numerator and denominator with the conjugate of the denominator, in a way similar to when rationalizing an expression:

$$\begin{aligned} \frac { 4+3i }{ 5+2i } &=\frac { 4+3i }{ 5+2i } \cdot \frac { 5-2i }{ 5-2i } \\ &=\frac { (4+3i)(5-2i) }{ { 5 }^{ 2 }+{ 2 }^{ 2 } } \\ &=\frac { 20-8i+15i-6{ i }^{ 2 } }{ 29 } \\ &=\frac { 26 }{ 29 } +\frac { 7 }{ 29 } i\\\\ \Rightarrow a&=\frac { 26 }{ 29 }, b=\frac { 7 }{ 29 }. \ _\square \end{aligned}$$

Let's look at more examples to strengthen our understanding.

Perform the necessary operation to put$\overline { \left( \frac { 2-3i }{ 4+5i } \right) \left( \frac { 4-i }{ 1-3i } \right) }$ to $a+bi \,(a,b \in \mathbb{R})$ form.

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Performing the necessary operations, and using the properties of complex numbers and their conjugates, we have

$$\begin{aligned} \overline { \left( \frac { 2-3i }{ 4+5i } \right) \left( \frac { 4-i }{ 1-3i } \right) } &=\overline { \left( \frac { 2-3i }{ 4+5i } \right) } \cdot \overline { \left( \frac { 4-i }{ 1-3i } \right) } \\\\ &=\frac { \overline { 2-3i } }{ \overline { 4+5i } } \cdot \frac { \overline { 4-i } }{ \overline { 1-3i } }\\\\ &=\frac { 2+3i }{ 4-5i } .\frac { 4+i }{ 1+3i } \\\\ &=\frac { 5+14i }{ 19+7i } . \end{aligned}$$

Use the rationalizing factor $19-7i$ to simplify:

$$\frac { 5+14i }{ 19+7i } \cdot \frac { 19-7i }{ 19-7i } =\frac { 193 }{ 410 } - \frac { 231 }{ 410 } i. \ _\square$$

Perform the necessary operation to put$\frac { -3x }{ 1-5xi } +\frac { 3i }{ 3+i }$ to $a+bi \,(a,b \in \mathbb{R})$ form.

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Rationalizing each term and summing up common terms, we have

$$\begin{aligned} \frac { -3x }{ 1-5xi } +\frac { 3i }{ 3+i } &=\frac { -3x }{ 1-5xi } \cdot \frac { 1+5xi }{ 1+5xi } +\frac { 3i }{ 3+i } \cdot \frac { 3-i }{ 3-i } \\ &= \left( \frac { -3x-15{ x }^{ 2 }i }{ 1+25{ x }^{ 2 } } \right) +\left( \frac { 9i+3 }{ 10 } \right) \\ &= \left( \frac { -3x }{ 1+25{ x }^{ 2 } } -\frac { 15{ x }^{ 2 } }{ 1+25{ x }^{ 2 } } i \right) +\left( \frac { 9 }{ 10 } i+\frac { 3 }{ 10 } \right) \\ &=\left( \frac { -3x }{ 1+25{ x }^{ 2 } } +\frac { 3 }{ 10 } \right) +\left( \frac { -15{ x }^{ 2 } }{ 1+25{ x }^{ 2 } } i+\frac { 9 }{ 10 } i \right) \\ &=\frac { -30x+3+75{ x }^{ 2 } }{ 10+250{ x }^{ 2 } } +\left( \frac { -150{ x }^{ 2 }+9+225{ x }^{ 2 } }{ 10+250{ x }^{ 2 } } \right) i. \ _\square \end{aligned}$$

Given that $x=5-i$ is a root of $f(x)=x^3-8x^2+6x+52,$ factor $f(x)$ completely.

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By the complex conjugate root theorem, we know that $x=5+i$ is also a root of $f(x).$ Hence,

$$\begin{aligned} \big(x-(5-i)\big)\big(x-(5+i)\big) &= \big((x-5)+i\big)\big((x-5)-i\big) \\ &=x^2-10x+26\end{aligned}$$

is a real factor of $f(x).$ We can divide $f(x)$ by this factor to obtain

$$\frac{x^3-8x^2+6x+52}{x^2-10x+26}=x+2.$$

Thus, $f(x)$ can be factored as

$$f(x)=(x-5+i)(x-5-i)(x+2). \ _\square$$

Find the cubic polynomial that has roots $5$ and $3+i.$

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According to the complex conjugate root theorem, $3-i$ which is the conjugate of $3+i$ is also a root of the polynomial. Hence, let $f(x)$ be the cubic polynomial with roots $3+i,$ $3-i,$ and $5,$ then

$$\begin{aligned} f(x) &= (x-5)\big(x-(3+i)\big)\big(x-(3-i)\big) \\ &= (x-5)\big((x-3)-i\big)\big((x-3)+i\big) \\ &= (x-5)\big(x^2-6x+9-i^2\big) \\ &= (x-5)\big(x^2-6x+10\big) \\ &= x^3-11x^2+40x-50. \ _\square \end{aligned}$$

If $p$ and $q$ are real numbers and $2+\sqrt{3}i$ is a root of $x^2+px+q=0,$ what are the values of $p$ and $q?$

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Since the coefficients of the quadratic equation are all real numbers, $2-\sqrt{3}i$ which is the conjugate of $2+\sqrt{3}i$ is also a root of the quadratic equation. Thus, by Vieta's formular

$$-p=\left(2+\sqrt{3}i\right)+\left(2-\sqrt{3}i\right),\quad q=\left(2+\sqrt{3}i\right)\left(2-\sqrt{3}i\right).$$

Therefore, $p=-4$ and $q=7. \ _\square$

Prove that if $a+bi \ (b \neq 0)$ is a root of $x^2+px+q=0$ and $a, b, p, q \in \mathbb{R},$ then $a-bi$ is also a root of the quadratic equation.

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Since $a+bi$ is a root of the quadratic equation, it must be true that

$$(a+bi)^2+p(a+bi)+q=0.$$

If we rewrite above equation, we obtain

$$\big(a^2-b^2+pa+q\big)+(2ab+pb)i=0.$$

Since $a, b, p, q \in \mathbb{R},$ we have

$$a^2-b^2+pa+q=0, \quad 2ab+pb=0. \qquad (1)$$

Now, if we substitute $a-bi$ into $x^2+px+q,$ then we obtain

$$(a-bi)^2+p(a-bi)+q=\big(a^2-b^2+pa+q\big)-(2ab+pb)i,$$

which is zero by $(1).$

Therefore, it must be true that $a-bi$ is also a root of the quadratic equation. $\ _\square$

Can the two complex numbers $\sin x+i\cos 2x$ and $\cos x-i\sin 2x$ be the conjugates of each other? If so, what is the possible real value for $x?$

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Let $\cos x-i\sin 2x$ be the conjugate of $\sin x+i\cos 2x,$ then we have $$\begin{aligned} \overline{\sin x+i\cos 2x} &= \cos x-i\sin 2x \\ \Rightarrow \sin x-i\cos 2x &= \cos x-i\sin 2x, \end{aligned}$$ which means $$\sin x=\cos x \text{ and } \cos 2x=\sin 2x$$ since the values of sine or cosine functions are real numbers. So we can rewrite above equations as follows: $$\tan x=1 \text{ and } \tan 2x =1.$$ Observe that these two equations cannot hold simultaneously, then the two complex numbers in the problem cannot be the conjugates of each other for any real value $x. \ _\square$

If $\alpha^2=3-4i,$ what is the value of $\alpha \overline{\alpha}?$

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Since $\alpha^2=3-4i,$ we have $$\left(\overline{\alpha}\right)^2=\overline{\alpha^2}=3+4i.$$ Hence, $$\begin{aligned} \left(\alpha \overline{\alpha}\right)^2 &= \alpha^2 \left(\overline{\alpha}\right)^2\\&=(3-4i)(3+4i)\\ &= 25 \\ \Rightarrow \alpha \overline{\alpha} &= \pm 5. \end{aligned}$$ Observe that if $\alpha=p+qi \ (p, q \in \mathbb{R})$ and $\overline{\alpha}=p-qi ,$ then $\alpha \overline{\alpha}=p^2+q^2 \geq 0.$ Then $$\alpha \overline{\alpha}=5. \ _\square$$

For a non-real complex number $\alpha,$ if $\alpha+\frac{1}{\alpha}$ is a real number, what is the value of $\alpha \overline{\alpha}?$

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Since $\alpha+\frac{1}{\alpha}$ is a real number, we have $$\alpha+\frac{1}{\alpha} = \overline{\left(\alpha+\frac{1}{\alpha}\right)}=\overline{\alpha}+\frac{1}{\overline{\alpha}}.$$ Therefore, we obtain $$\begin{aligned} \left(\alpha-\overline{\alpha}\right)+\left(\frac{1}{\alpha}-\frac{1}{\overline{\alpha}}\right) &= 0 \\ \left(\alpha-\overline{\alpha}\right)\left(1-\frac{1}{\alpha \overline{\alpha}}\right) &= 0. \end{aligned}$$ Since $\alpha$ is a non-real number, $\alpha \neq \overline{\alpha}.$ Thus, $$1-\frac{1}{\alpha \overline{\alpha}}=0,$$ which implies $\alpha \overline{\alpha}=1. \ _\square$

If a complex number $z=a+bi \ (a > 0, b > 0)$ satisfies $z^2+\overline{z}=0,$ how many positive integers $n$ less than 100 are there that make $z^n$ an integer?

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Since $z^2+\overline{z}=0,$ we have $$\begin{aligned} z^2+\overline{z} &= (a+bi)^2+(a-bi) \\ &= (a^2-b^2+a)+(2ab-b)i=0. \end{aligned}$$ Hence, $$\begin{aligned} a^2-b^2+a &= 0 \qquad (1) \\ 2ab-b &= 0 \\ \Rightarrow b(2a-1) &=0. \qquad (2)\end{aligned}$$ Since $b > 0,$ we obtain $a=\frac{1}{2}$ from $(2),$ and by substituting this into $(1)$ we have $b^2=\frac{3}{4}$ or $b=\frac{\sqrt{3}}{2}$ since $b > 0.$ Therefore, $$z=\frac{1+\sqrt{3}i}{2}.$$ Now, observe that $$\begin{aligned} z^2 &= \frac{-1+\sqrt{3}i}{2} \\ z^3 &= zz^2=\frac{1+\sqrt{3}i}{2} \cdot \frac{-1+\sqrt{3}i}{2}=-1 \\ z^4 &= zz^3=\frac{1+\sqrt{3}i}{2} \cdot (-1)=\frac{-1-\sqrt{3}i}{2} \\ z^5 &= z^2z^3=\frac{-1+\sqrt{3}i}{2} \cdot (-1)=\frac{1-\sqrt{3}i}{2} \\ z^6 &= \big(z^3\big)^2=1 \\ &\vdots, \end{aligned}$$ then $n$ must be a multiple of 3 to make $z^n$ an integer.

Thus, there are 33 positive integers less then 100 that make $z^n$ an integer. $\ _\square$