Question

5. (i) Find the product of the two complex numbers (3 - 2i) and (2 + 5i) and hence show that the modulus of
the product is equal to the product of their moduli​

Answer 1

Answer:

attached above mate

Step-by-step explanation:

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Answer 2

Answer:

The product of the given complex numbers is found to be $16+11i$  and the product of the modulus of two complex numbers is proven equal to the modulus of their product.

Step-by-step explanation:

The given two complex numbers to us are:

$3-2i$  and  $2+5i$

The product of the two complex numbers will be:

$X=(3-2i)(2+5i)$

$X=3(2)+3(5i)-2i(2)-2i(5i)$

Simplifying it, we get:

$X=6+15i-4i-10i^2$

Using the identity of $i^2=-1$, we get:

$X=6+11i+10$

or we can say:

$X=16+11i$

So, the product of the two complex numbers is found to be: $16+11i$.

Now, finding the modulus of X, we get:

$|X|=\sqrt{(16)^2+(11)^2}$

Simplifying it, we get:

$|X|=\sqrt{256+121}$

or we can say:

$|X|=\sqrt {377}$                              ...(i)

Now, finding the product of modulus of the given complex numbers, we get:

$|X_1|.|X_2|=(\sqrt{(3)^2+(-2)^2})(\sqrt{(2)^2+(5)^2})$

Simplifying it, we get:

$|X_1|.|X_2|=(\sqrt{9+4})(\sqrt{4+25})$

$|X_1|.|X_2|=(\sqrt{13})(\sqrt{29})$

or we can say:

$|X_1|.|X_2|=\sqrt{29\times 13}$

After multiplication, we get:

$|X_1|.|X_2|=\sqrt{377}$                      ...(ii)

Hence (i) = (ii),

Thus, the product of modulus of two complex numbers is equal to the modulus of the product of two complex numbers.