Корені з комплексного числа. Спряжені числа

The complex conjugate

In the previous explanation we described the interchangeability of $i$ and $-i$. Now, let's do it!

We replace every $i$ by $-i$: $7i$ becomes $-7i$, $2+3i$ becomes $2-3i$, and in general $a+bi$ becomes $a-bi$.

We say $a-bi$ is the complex conjugate of $a+bi$ and we write this as $\overline{a+bi} = a-bi$.

The usual arithmetic still works: $\overline{a+bi}+\overline{c+di} = (a+c)-(b+d)i = \overline{(a+bi)+(c+di)}$ $\overline{a+bi}\cdot\overline{c+di} = (ac-bd)-(ad+bc)i = \overline{(a+bi)(c+di)}$

We can do exactly the same things with conjugated numbers!

Their strength comes when we combine them with unconjugated numbers. Let $z$ be the complex number $a+bi$ and let $\overline{z}$ be its complex conjugate. These are a some powerful identities:

$Re(z) = a = \frac{z+\overline{z}}{2}$ $Im(z) = b = \frac{z-\overline{z}}{2}$ $|z|^2 = a^2+b^2 = z\overline{z}$

Now, let's use them to solve some problems!

Division of a complex number

Problem 1: Write $\frac{3+5i}{1+i}$ in Cartesian form.

We want the complex number in the denominator to disappear. We know $z\overline{z} = |z|^2$ is a real number, so we multiply both the numerator and the denominator by $1-i$. We get

$$\frac{3+5i}{1+i} = \frac{(3+5i)(1-i)}{(1+i)(1-i)} = \frac{8+2i}{2} = 4+i.$$

Finding solutions to a polynomial equation

Problem 2: Find the solutions of $x^2-2x+2 = 0$.

After a lot of guesses we begin to suspect that this equation has no real solutions. So we try the complex solution $a+bi$. If this is a solution, $a-bi$ must be a solution as well. We can write

$$0 = x^2-2x+2 = (x-(a+bi))(x-(a-bi)) = x-2ax+a^2+b^2$$

and see that $a$ and $b$ should equal $1$. We check the solutions $1+i$ and $1-i$ and observe they are correct!

The complex conjugate of the complex number $z=a+bi$, with $a$ and $b$ real, is $\overline z=a-bi$. For example, the complex conjugate of $8+6i$ is $8-6i$.

There are several important properties with complex conjugates. Suppose $z$ and $w$ are complex numbers. Then:

(1) $\overline{z+w}=\overline z+\overline w$

(2) $\overline{z-w}=\overline z-\overline w$

1. $\overline{z\cdot w}=\overline z\cdot\overline w$

I will prove (1) and leave (2) and (3) as an exercise.

Proof for (1). Let $z=a+bi$ and $w=c+di$. Then, $$\begin{align*}\overline{z+w}&=\overline{(a+bi)+(c+di)}\\&=\overline{(a+c)+(b+d)i}\\&=(a+c)-(b+d)i\\&=(a-bi)+(c-di),\end{align*}$$ which is exactly $\overline z+\overline w$.

Here's another useful property: $$|z|^2=z\overline z.$$ You can prove this using the same idea as above. ($|z|$ is the distance from $z$ to the origin in the complex plane)

Problem: Simplify $\dfrac{1+i}{1-i}$.

We can use the above property. The important part is that when we multiply a complex number with its conjugate, we get a real number. So we can try $$\frac{1+i}{1-i}=\frac{1+i}{1-i}\cdot\frac{1+i}{1+i}=\frac{(1+i)^2}{2}=i.$$

Now you have the tools to prove the following:

$\overline{z/w}=\overline z/\overline w$ (when $w$ is nonzero, of course)

$\overline z=z$ if and only if $z$ is real

$\text{Real part of }z=\mathfrak{Re}(z)=\dfrac{z+\overline z}2$

$\text{Imaginary part of }z=\mathfrak{Im}(z)=\dfrac{z-\overline z}{2i}$

When we have an fraction with $i$ in both the numerator and the denominator, we want to transform the fraction back to standard form: $xi + y$, where $x$ and $y$ are real constants. This is an example of a fraction complex number: $(5)/(9 + 2i)$

The denominator, $(9 + 2i)$, has a conjugate, $(9 - 2i)$. It can be multiplied to its conjugate to form a difference of squares, consisting of only numeral terms and $i^2$ terms. The $i^2$ term can then be symplified to a $-1$ term, making all the terms constant numerals. Let's do this on the example fraction complex number:
$(5)/(9 + 2i)$ = $(5) * (9 - 2i)/(9 + 2i) * (9 - 2i)$ = $(45 - 10i)/(81 - 4i)$

Your answer is: $(45 - 10i)/(81 - 4i)$.

Congratulations!

(Video) Complex Conjugates

by mathman1024

Here's a video by mathman1024 explaining complex conjugates. A conjugate is when you change the middle of two like terms. So the conjugate of $a+b$ would be $a-b$. Multiplying conjugates gives you the difference of squares, $a^2-b^2$.

So complex conjugates are when there's an imaginary number involved. They are usually in the form: $$(a+bi)(a-bi)$$ To multiply them we can use the product of squares formula. $$(a)^2-(bi)^2\\ a^2-b^2i^2$$ Remember though, $i^2=-1$. $$a^2-b^2(-1)\\ a^2+b^2$$ As you see, multiplying complex conjugates removes the imaginary number and leaves a real number.

He then goes over some problems involving complex conjugates. The first is: $(7+2i)(7-2i)$. $$(7+2i)(7-2i)\\ 7^2+2^2\\ 49+4\\ 53$$ He then does some more examples. Try them on your own and see if you get the same answers he does.

1. $(11-i)(11+i)$
2. $(-8+5i)(-8-5i)$

Any complex number, $z$, has a conjugate, $\bar{z}$ (sometimes written as $z^{*}$). If we have a complex number in rectangular form, $$\begin{align}z&=a+bi, \text{then,}\\ \bar{z}&=a-bi\end{align}$$

Both have the imaginary and real parts. Think of it as taking the conjugate of the denominator with rational numbers when you rationalize!

Put in a more general way, the complex conjugate of $z$ has the same real part, but opposite imaginary part.

$$\begin{align}Re[z]&=Re[\bar{z}] \\ Im[z]&=-Im[\bar{z}]\end{align}$$

This actually applies to a complex number written in any form. Let's examine this with polar form. $$\begin{align}z&=a+bi \\ &=\sqrt{a^{2}+b^{2}}e^{i\cdot \arctan\left(\frac ba\right)} \\ \\ \bar{z}&=a-bi \\ &=\sqrt{a^{2}+b^{2}}e^{i\cdot \arctan\left(-\frac ba\right)} \\ &=\sqrt{a^{2}+b^{2}}e^{-i\cdot\arctan\left(\frac ba\right)}\end{align}$$

As can be seen, in order to form the conjugate all we had to do was put a minus sign in front of the $i$.

It turns out that multiplying a complex number by its conjugate is an easy way of turning it into a real number. This stems from the property of $i^{2}=-1$. In fact, we can use the conjugate to simply and easily find the magnitude of any complex number. All we need to know is $|z|=\sqrt{z\bar{z}}$. Let's do two quick examples, one in rectangular form, and one in polar. $$z=a+bi \quad \bar{z}=a-bi$$ $$\begin{align}|z|&=\sqrt{z\bar{z}} \\ &=\sqrt{(a+bi)(a-bi)} \\ &=\sqrt{a^{2}-\cancel{abi}+\cancel{abi}-b^{2}i^{2}} \\ &=\sqrt{a^{2}+b^{2}}\end{align}$$ This is the correct magnitude for our complex number. Now let's do the polar example. $$z=Ae^{i\theta} \quad \bar{z}=Ae^{-i\theta}$$ $$\begin{align}|z|&=\sqrt{z\bar{z}}\\&=\sqrt{Ae^{i\theta}\cdot Ae^{-i\theta}}\\&=\sqrt{A^{2}e^{i\theta-i\theta}}\\&=\sqrt{A^{2}e^{0}}\\&=\sqrt{A^{2}}\\&=A\end{align}$$ This is the correct magnitude for our complex number.

The magnitude of $z=\frac{2}{5+7i}$ is as follows, $$\begin{align} |z|&=\sqrt{z\bar{z}} \\ &=\sqrt{\frac{2}{5+7i}\cdot \frac{2}{5-7i}} \\ &=\sqrt{\frac{4}{25+49}} \\ &=\sqrt{\frac{4}{74}} \\ &=\sqrt{\frac{2}{37}}\end{align}$$

Moving forward, the key things to remember are; the complex conjugate is by swapping the sign on all $i$'s, and the complex conjugate is useful to find the magnitude of a complex number

Complex Conjugates

We are familiar with regular conjugates, and complex conjugates are just a step further, involving imaginary numbers!

There are three main characteristics with complex conjugates:

• Opposite signs: the signs are opposite, so one conjugate has a positive sign and one conjugate has a negative sign

• Two like terms: the terms within the conjugates must be the same

• Product is a Sum of Squares: unlike regular conjugates, the product of complex conjugates is the sum of squares! The $i^2=-1$, which flips the sign.