Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

How to find the Euclidean norm of a complex number, like $10+i$ or $2-i$? Explain with clear, easy detail.

share|cite|improve this question
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:… – 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? – user27825 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
Please phrase requests as requests, not orders. Thank you. – Arturo Magidin Mar 28 '12 at 15:49

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.

share|cite|improve this answer
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
@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

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 $z$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.