Question

(2+3i) (2-3i) find in the form of a+bi​

Answer 1

Answer:

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

4-6i+6i-9i^2

6i will cancel each other

Value of i=under root - 1

So the value of i^2=-1

So we get

4+9=13

So ans in a+bi form is

13+0i

Answer 2

(2 + 3i) (2 - 3i) = 13 + 0i which is of the form of a + bi where a = 13 , b = 0

Given :

The expression (2 + 3i) (2 - 3i)

To find :

To express (2 + 3i) (2 - 3i) in the form of a + bi

Solution :

Step 1 of 3 :

Write down the given expression

The given expression is (2 + 3i) (2 - 3i)

Step 2 of 3 :

Simplify the given expression

$\displaystyle \sf{ (2 + 3i) (2 - 3i) }$

$\displaystyle \sf{ = {(2)}^{2} - {(3i)}^{2} }$

$\displaystyle \sf{ = 4 - 9 {i}^{2} }$

$\displaystyle \sf{ = 4 - \{ 9 \times ( - 1) \} }\: \: \: \bigg[ \: \because \:{i}^{2} = - 1 \bigg]$

$\displaystyle \sf{ = 4 + 9 }$

$\displaystyle \sf{ = 13 }$

$\displaystyle \sf{ = 13 + 0i }$

Step 3 of 3 :

Express in the form of a + bi

(2 + 3i) (2 - 3i) = 13 + 0i

which is of the form of a + bi where a = 13 , b = 0

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