How is $(2+i\sqrt{2}) \cdot (2-i\sqrt{2})$ calculated?

Question

What is $(2+i\sqrt{2}) \cdot (2-i\sqrt{2})$ ?

Answer:

(a) $4$

(b) $6$

(c) $8$

(d) $10$

(e) $12$

I calculate like this:

$(2+i\sqrt2),(2-i\sqrt2)\\(2+1.41421i),\;(2-1.41421i)\\3.41421i,\;0.58579i\\3.41421i+0.58579i\\4i$

Therefore, the answer is $4$.

But the correct answer is $6$.

How is it calculated correctly?

Answer 1

Use $(P+Q)(P-Q)=P^2-Q^2$ and $i^2=-1$

Then $$F=(2+i\sqrt{2})(2-i\sqrt{2})=4-2i^2=4+2=6$$

Answer 2

There is no need to approximate $\sqrt2$ by a decimal. You seem to have concluded $4i=4$ which is clearly incorrect. It's helpful to understand and apply $$ (a+ib)(a-ib)=a^2+b^2 $$

Answer 3

Use identity: $\color{blue}{(a+b)(a-b)=a^2-b^2}$ and $i=\sqrt{-1}$ or $\color{blue}{i^2=-1}$ $$(2+i\sqrt2)(2-i\sqrt2)=(2)^2-(i\sqrt2)^2=4-2i^2=4-2(-1)=4+2=6$$

Answer 4

In general, multiplication is distributive over addition,

so $(a+b)\cdot(c+d)=a\cdot c+a\cdot d+b\cdot c+b\cdot d$.

In this particular case,

$\require{enclose}(2+i\sqrt2)\cdot(2-i\sqrt2)=2\cdot2{-\enclose{downdiagonalstrike}{2\cdot i\sqrt2}+\enclose{downdiagonalstrike}{2\cdot i\sqrt2}}-i^2\sqrt2^2=4-(-1)2=6.$

Answer 5

By distributivity of multiplication over addition, we have

$$\color{red}{(2+i\sqrt{2})}(2-i\sqrt{2})=\color{red}{(2+i\sqrt{2})}2-\color{red}{(2+i\sqrt{2})}i\sqrt{2}.$$

Distributing again gives

$$(2+i\sqrt{2})\color{blue}{2}-(2+i\sqrt{2})\color{green}{i\sqrt{2}}=2\times\color{blue}{2}+i\sqrt{2}\times\color{blue}{2}-2\times\color{green}{i\sqrt{2}}-(i\sqrt{2})\times\color{green}{i\sqrt{2}},$$

so that we have

$$\begin{align} (2+i\sqrt{2})(2-i\sqrt{2})&=4+\enclose{downdiagonalstrike}{{i2\sqrt{2}}}-\enclose{downdiagonalstrike}{{i2\sqrt{2}}}-(i\sqrt{2})^2\\ &=4-(-2)\\ &=6. \end{align}$$