How to find the Euclidean norm of a complex number?

Question

I would like to know how to find the Euclidean norm of a complex number, like $10+i$ or $2-i$?

I would appreciate a clear and easy explanation with the necessary details.

Step 1: Look up the definition. Step 2: perform the simple calculation given in the definition. — Chris Eagle, Mar 28 '12 at 13:30
$z=x+iy$ and read here: http://en.wikipedia.org/wiki/Norm_(mathematics)#Euclidean_norm_of_a_complex_number — Salech Alhasov, Mar 28 '12 at 13:32
|1+i|e=(1+i^2)^.5=1-1=0, but the answer is supposed to be 2. So we're multiplying by conjugate, why? — , Mar 28 '12 at 15:43
Very confused by what you wrote there. What is e? How did you get to (1+i^2)^.5 ? As you said, you need to multiply by the conjugate. — Ted, Mar 28 '12 at 15:48

score 6 · Answer 1 · edited Mar 28 '12 at 18:19

6

For $x + iy \in \mathbb C$ the Euclidean norm is defined as $\| x + iy \| := \sqrt{x^2 + y^2}$.

Now you need to fill in the numbers and compute.

edited Mar 28 '12 at 18:19

answered Mar 28 '12 at 13:33

Rudy the Reindeer

1

Ping me if you can't do it. But let me see some calculations first. – Rudy the Reindeer Mar 28 '12 at 13:33
|1+i|=(1+i^2)^.5=1-1=0 Why are we multiplying by the conjugate? – nnb Mar 28 '12 at 18:03
1

@nnb: You are computing incorrectly: $1+i = 1+1i$, so $x=y=1$ in this case; the formula says $|1+i|=|1+1i| = \sqrt{1^2+1^2} = \sqrt{2}$. Multiplying by the conjugate accomplishes the same thing: $(1+i)(1-i) = 1^2-(i)^2 = 1^2-(-1) = 2$. More generally,$$(x+iy)(x-iy) = x^2 - (iy)^2 = x^2 -(-1)y^2 = x^2+y^2.$$ – Arturo Magidin Mar 28 '12 at 18:36
@nnb What Arturo said. Thanks, Arturo. – Rudy the Reindeer Mar 28 '12 at 18:50

score 4 · Answer 2 · edited Mar 29 '12 at 00:43

I am adding an answer especially to clear up OP's confusion about multiplying conjugates and its relation with the Euclidean norm.

Let $a+b i$ be a complex number. Note that its euclidean norm, which I'll denote by $\| \cdot \|_e$, is given by $$\|a+bi\|_e=\sqrt {a^2+b^2}$$

Now what is the conjugate of $a+bi$? We know that it is given by $\overline{a+bi}$ which equals $a-bi$.

So, $$\begin{align}(a+bi)(a-bi)&=a^2-abi+abi-b^2i^2\\ &=a^2-\not{abi}+\not{abi}+b^2 ~~~~\mbox{as $i^2=-1$}\\ &=\|a+bi\|_e^2\end{align}$$

So, to get its euclidean norm, it helps to multiply by its conjugate and take its positive square root.

That is, for a complex number $z$, we have that $$\|z\|_e=\sqrt{z\bar z}$$ where $\bar z$ denotes the conjugate of $z$.

2 Answers2