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:

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

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

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

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

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

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,

\[\begin{align} \big(x-(5-i)\big)\big(x-(5+i)\big) &= \big((x-5)+i\big)\big((x-5)-i\big) \\ &=x^2-10x+26\end{align}\]

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

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

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 \[\begin{align} \overline{\sin x+i\cos 2x} &= \cos x-i\sin 2x \\ \Rightarrow \sin x-i\cos 2x &= \cos x-i\sin 2x, \end{align} \] 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, \[\begin{align} \left(\alpha \overline{\alpha}\right)^2 &= \alpha^2 \left(\overline{\alpha}\right)^2\\&=(3-4i)(3+4i)\\ &= 25 \\ \Rightarrow \alpha \overline{\alpha} &= \pm 5. \end{align}\] 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 \[\begin{align} \left(\alpha-\overline{\alpha}\right)+\left(\frac{1}{\alpha}-\frac{1}{\overline{\alpha}}\right) &= 0 \\ \left(\alpha-\overline{\alpha}\right)\left(1-\frac{1}{\alpha \overline{\alpha}}\right) &= 0. \end{align}\] 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 \[\begin{align} z^2+\overline{z} &= (a+bi)^2+(a-bi) \\ &= (a^2-b^2+a)+(2ab-b)i=0. \end{align}\] Hence, \[\begin{align} a^2-b^2+a &= 0 \qquad (1) \\ 2ab-b &= 0 \\ \Rightarrow b(2a-1) &=0. \qquad (2)\end{align}\] 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 \[\begin{align} z^2 &= \frac{-1+\sqrt{3}i}{2} \\ z^3 &= zz^2=\frac{1+\sqrt{3}i}{2} \cdot \frac{-1+\sqrt{3}i}{2}=-1 \\ z^4 &= zz^3=\frac{1+\sqrt{3}i}{2} \cdot (-1)=\frac{-1-\sqrt{3}i}{2} \\ z^5 &= z^2z^3=\frac{-1+\sqrt{3}i}{2} \cdot (-1)=\frac{1-\sqrt{3}i}{2} \\ z^6 &= \big(z^3\big)^2=1 \\ &\vdots, \end{align}\] 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. \(\ _\square\)