Question

Simply it
$\sf\dfrac{1 + 3i}{1 - 2i}$

Answer 1

1+3i/1–2i

=(1+3i)(1+2i)/(1–2i)(1+2i) ; we multiply both numerator and denominator with (1+2i) so that we can get a denominator free from i

={1+3i+2i+(3i×2i)}/{(1)^2–(2i)^2} ; since (a+b)(a-b)=a^2-b^2

=(1+5i+6i^2)/(1–4i^2)

={1+5i+6(-1)}/{1–4(-1)} ; since i^2=-1

=(1+5i-6)/(1+4)

=5i-5/5

=5(i-1)/5

=i-1 ; or

=(√-1) - 1

Answer 2

EXPLANATION.

Simplify,

⇒ 1 + 3i/1 - 2i.

As we know that,

We can rationalize the equation, we get.

⇒ 1 + 3i/1 - 2i X 1 + 2i/1 + 2i.

⇒ (1 + 3i)(1 + 2i)/(1 - 2i)(1 + 2i).

⇒ (1 + 2i + 3i + 6i²)/(1 - 4i²).

As we know that,

⇒ i² = -1.

Put the values in the equation, we get.

⇒ (1 + 5i + 6(-1))/(1 - 4(-1)).

⇒ (5i - 6 + 1)/(1 + 4).

⇒ (5i - 5)/5.

⇒ 5(i - 1)/5.

⇒ i - 1.

                                                                                                                             

MORE INFORMATION.

Properties of conjugate complex number.

if z = a + ib.

$(1) = \sf (\bar{z}) = z$

$\sf (2) = z + \bar{z} = 2a = 2 \ Re (z) = purely \ real.$

$\sf (3) = z - \bar{z} = 2ib = 2i \ Im (z) = purely \ imaginary.$

$\sf (4) = z\bar{z} = a^{2} + b^{2} = |z|^{2} .$

$\sf (5) = z + \bar{z} = 0 \ Or \ z = -\bar{z} = z = 0 \ or \ z \ is \ purely \ imaginary.$

$\sf (6) = z = \bar{z} = z \ is \ purely \ real.$