Question

Z1= 4+3i Z2= 2-3i find Z1/Z2

complex numbers​

Answer 1

Answer:

$\frac{-1+18i}{13}$

Step-by-step explanation:

$\frac{z_1}{z_2} = \frac{4+3i}{2-3i} = \frac{4+3i}{2-3i}.\frac{2+3i}{2+3i} = \frac{8-9+18i}{4+9}\\=\frac{-1+18i}{13}$

Answer 2

Answer:

The value of $\frac{Z_1}{Z_2}$ =  $\frac{-1+18i}{13}$

Step-by-step explanation:

Given,

$Z_1$= 4+3i  and  $Z_2$= 2-3i

Recall the concepts

1. i² = -1

2. (a+b)(a-b) = a² - b²

Required to find $\frac{Z_1}{Z_2}$

Solution

$\frac{Z_1}{Z_2}$ = $\frac{4+3i}{2-3i}$

To simplify the above fraction, we should multiply the numerator and denominator with the complex conjugate of the denominator

The complex conjugate of the denominator = 2+3i

$\frac{4+3i}{2-3i}$ = $\frac{4+3i}{2-3i} X\frac{2+3i}{2+3i}$

= $\frac{(4+3i)(2+3i)}{(2-3i)(2+3i)}$

= $\frac{(8+12i+6i+(3i)^2)}{(2^2 - (3i)^2}$((a+b)(a-b) = a²-b²)

=$\frac{8+18i-9}{4 - (-9)}$( i² = -1)

= $\frac{-1+18i}{13}$

$\frac{Z_1}{Z_2}$ =  $\frac{-1+18i}{13}$

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