Question

(3+2i)(3-2i)
Solve the above expression

Ch - Complex Numbers
Class 11

Answer 1

Answer:

$(3+2i)(3-2i) = 13$

Step-by-step explanation:

We have been given the statement:-

$(3+2i)(3-2i)$

Using the identity:-

$(a+b)(a-b) = a^{2} - b^{2}$

$= (3+2i)(3-2i) = (3)^{2} - (2i)^{2}$

$= (3)^{2} - (2i)^{2}$

$= 9 - 4i^{2}$

Value of $i^{2} = -1$

$= 9 - 4(-1)$

$= 9 + 4$

$= 13$

So, the answer is 13 !

Answer 2

Answer:  

$\left(3+2i\right)\left(3-2i\right)=13$  

Step-by-step explanation:  

We are given a complex expression;

$\left(3+2i\right)\left(3-2i\right)$  

 

$\mathrm{Let's\:apply\:complex\:arithmetic\:rule}:\quad \left(a+bi\right)\left(a-bi\right)=a^2+b^2$  

$a=3,\:b=2$  

$=3^2+2^2$  

 

$\mathrm{Refine}$  

$=13$