# What is the absolute value of the follwing: z = (1+i)/(2-3i)

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### 2 Answers

z =* (1 + i)/(2- 3i)*

let z = a + bi

(a + bi )(2 - 3i) = 1 + i

2a + 2bi - 3ai + 3b = 1 + i

This creates a system of two equations by combining real and imaginary components:

2a + 3b = 1

2b - 3a = 1

Solving,

a = -1/13, b = 5/13So, * (2-3i)/(2+i) *= -1/13(1 + 5)i

r = 0.39 ,theta = 101.3° in polar coordinates.

Thus the absolute value is 0.39

z = (1+i)/(2-3i)

First we will rewrite z in the standard form of the complex number

which is z = a+ bi

To do that we need to multiply and divide by the denominator inverse (2+3i)

==> z = (1+i)(2+3i)/ (2-3i)(2+3i)

(1+ i)(2+3i) = (2 + 3i + 2i + 3i^2

But i^2 = -1

==> (1+i)(2+3i) = 2 + 5i - 3 = -1 + 5i

Now the denominator:

(2-3i)(2+3i) = ( 4 - 9i^2 = 4 + 9 = 13

Now let us rewrtie:

z = (-1+5i) / 13

= (-1/13 ) + (5/13) i

==> l z l = sqrt(-1/13)^2 +(5/13)^2

= sqrt26 / 13

**Then the absolute value for z is = sqrt26/ 13**