# How to calculate $(3+4i)\cdot(1+i)$

I have recently read an article on imaginary numbers. It was very interesting, but left me with the above question. It had the answer in the question, it was $-1+7i$. But how do I calculate this?

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Try using ordinary multiplication, replacing $i^2$ with $-1$ when needed. – enzotib Sep 22 '12 at 7:47
Think of $3+4i$ as a pair of numbers $(3,4)$ with the multiplication rule $(a,b) \cdot (c,d) = (ac-bd, ad+bc)$. Then $(3+4i)\cdot (1+i) = (3,4) \cdot (1,1) = (-1,7) = -1+7i$. In particular, $i = (0,1)$ which gives $i^2 = (0,1)^2 = (-1,0) = -1$. – copper.hat Sep 22 '12 at 8:10

$(3+4i)\cdot(1+i)$

$= 3+4i+3i+4\cdot i^{2}$

$= 3+7i + 4\cdot i^{2}$

$i$ can be represented as ${\sqrt{-1}}$

So, ${i^{2} = -1}$

So, the above can be writen as :

$3+7i-4 = -1+7i$

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Thanks a lot! Just didn't understand how i could be used in maths until today. – imulsion Sep 22 '12 at 8:03
@enzotib: Thanks – Sumit Bhowmick Sep 23 '12 at 4:27
Argh! That is a horribly non-standard use of $\Longrightarrow$. Please don't do that when you just mean $=$. – Henning Makholm Sep 23 '12 at 13:46
@HenningMakholm: I have edited – Sumit Bhowmick Sep 28 '12 at 4:55

For complex numbers many of the rules for real numbers also hold. In particular the distributive law, so you can multiply out: \begin{align*} (3+4i)\cdot (1+i)&= 3\cdot (1+i) + 4i\cdot (1+i)\\ &= 3\cdot 1 + 3\cdot i + 4i\cdot 1+ 4i\cdot i\\ &= 3 + 3i + 4i + 4\cdot (-1)\\ &= -1 + 7i \end{align*}

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I presume you meant 'rules for real numbers also hold'? – copper.hat Sep 22 '12 at 8:11
Yes, thanks. Edited. – Julian Kuelshammer Sep 22 '12 at 8:24

This is basic binomial multiplication. Using the fact that $(a + b)\cdot(c + d) = ac + ad + bc + bd$, we have,\begin{align}(3 + 4i)\cdot (1 + i) & = &3 + 3i + 4i + 4i^2 \\ & = & 4i^2 + 7i + 3\\ & = & 4(-1) + 7i + 3\\ & = & -4 + 7i + 3\\&=&7i - 1 \text{ which could be written as}-1+7i \end{align}*Important Note: In step 3, the fact that $i^2 = -1$ is used.*

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