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According to Wolfram Alpha, Re(1/(2+i))=2/5.

How did it calculate that?

amWhy
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Hal
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    Multiply top and bottom by $2-i$. – icurays1 Sep 22 '14 at 14:20
  • @icurays1 If you didn't know the solution was 2/5, would you know to do that? If so, how did you know that would give you the solution? – Hal Sep 22 '14 at 14:24
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    @Hal: Because that's the standard method for computing division into 'standard' form. –  Sep 22 '14 at 14:25
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    Your goal is to get all the $i$'s in the numerator. So you need to somehow get rid of the $i$ in the denominator. The way you can do that is if you multiply the complex number in the denominator with it's conjugate. Conjugate in this case happened to be $2-i$ – Sheheryar Zaidi Sep 22 '14 at 14:26
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    @Hal it seems mysterious at first, but the trick is simple: we hate nasty things like complex numbers in the denominator. The simplest number $w$ so that $zw$ is real is $\bar{z}$, so we multiply top and bottom by it. – icurays1 Sep 22 '14 at 14:41
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    Why did anyone down-vote this question? It seems perfectly fine to me. – Trevor J Richards Sep 22 '14 at 14:49
  • @Hal The reason why it works is that $z\cdot \overline{z} = |z|^2$ (where $\overline{z}$ is the complex conjugate of $z$, i.e. the change of sign of the imaginary part), and $|z|$ is always real. Therefore we can always make the denominator real by multiplying by the conjugate. – Eff Sep 22 '14 at 14:52

8 Answers8

7

$Re(z) = \frac 12 (z + \bar z)$

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

mercio
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$$\frac{1(2-i)}{(2+i)(2-i)} = \frac{2-i}{5} = \underbrace{\frac{2}{5}}_\text{real part} + i\cdot \overbrace{(-\frac{1}{5})}^\text{imaginary part}$$

Sheheryar Zaidi
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5

Multiply numerator and denominator by $2-i$:

$$\frac 1{2+i}\cdot \frac{2-i}{2-i} = \frac{2-i}{4-(-1)} = \frac{2-i}{5} = \frac 25 - \frac 15\cdot i$$

Edit for more explanation as to why this strategy works.

We multiply numerator and denominator by the conjugate of $2+ i$, which is $2+i$, to remove the imaginary number from the denominator. Doing so gives us a real-valued denominator: $$(2 + i)(2-i) = 2^2 - (i^2) = 4 - (-1) = 5.$$ Having multiplied by $\frac{2-i}{2-i}=1$, that transforms the numerator to $1\cdot (2-i) = 2- i$. Then all that remained was to transform the fraction $\dfrac{2-i}{5}$ into its real and imaginary components.

The same strategy works for any fraction of the form $\dfrac{z_1}{z_2},\,$ where $z_1, z_2 \in \mathbb C$ and $z_2 = a + bi$ where $a, b \in \mathbb R, b\neq 0$. We simply multiply numerator and denominator by $\overline{z_2}$. The denominator is then $z_2\cdot \overline{z_2} = |z| = a^2 + b^2$

amWhy
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4

The unwritten goal for all of these answers is to write the complex number in the form $$(\textrm{real number}) + (\textrm{another real number})i$$ The first real number is the real part and the second real number is the imaginary part.

So if you have a complex number $a+bi$ (with $a$ and $b$ real) as the denominator of a fraction, you can multiply top and bottom of the fraction by the so-called conjugate $a-bi$ to get rid of that. This works because it is always true that $(a+bi)(a-bi) = a^2+b^2$ is real.

Then you end up with a complex number in the numerator, which is easier to deal with. For example, suppose we wanted to find the real and imaginary parts of $\frac{1+3i}{2+i}$. Then we would write $$\frac{1+3i}{2+i} = \frac{1+3i}{2+i}\cdot\left(\frac{2-i}{2-i}\right)$$ $$=\frac{(1+3i)(2-i)}{(2+i)(2-i)}$$ $$=\frac{2-i +6i+3}{4+1}$$ $$= \frac{5+5i}{5}$$ $$ = \left(\frac55\right) + \left(\frac55\right)i$$ $$= 1+i$$ So in this case, the real part is $1$ and the imaginary part also happens to be $1$.

MPW
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3

$$\frac{1}{2+i}=\frac{2-i}{(2+i)(2-i)}=\frac{2-i}{5}=\frac{2}{5}-\frac{1}{5}i$$

J.R.
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$$ \frac{1}{2+i} = \frac{2-i}{(2+i)(2-i)} $$

BaronVT
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1

since $\frac{1}{2+i}=\frac{2-i}{4+1}=\frac{2-i}{5}$

Dr. Sonnhard Graubner
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$\frac{1}{a+bi}=\frac{1}{a+bi}\frac{a-bi}{a-bi}=\frac{a}{a^2+b^2}+i\cdot bla$

$a=2,b=1\implies \frac{a}{a^2+b^2}=\frac{2}{2^2+1^2}=\frac{2}{5}$

Nikolaj-K
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