1. (a) Find conjugate and magnitude of $$ z=\frac{(3+2 i)(1-2 i)}{4+3 i} $$ (To be solved in elass) (b) If $$ = $$ is a complex number then $$ |z| \geq 0 $$ and $$ |z|=0 $$ iff $$ z=0 $$ (To be solved in class) $$ z=3(1-2 i)+2 i(1-2 i) $$

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(a) To find the conjugate and magnitude of the complex number $$ z=\frac{(3+2i)(1-2i)}{4+3i} $$, we first need to simplify the expression.

Given:
$$ z=\frac{(3+2i)(1-2i)}{4+3i} $$

To simplify the expression, we can multiply the numerator and denominator by the conjugate of the denominator to eliminate the complex number in the denominator.

The conjugate of $$ 4+3i $$ is $$ 4-3i $$.

Multiplying the numerator and denominator by $$ 4-3i $$, we get:

$$ z=\frac{(3+2i)(1-2i)(4-3i)}{(4+3i)(4-3i)} $$

Now, we can simplify the expression further:

$$ z=\frac{(3+2i)(1-2i)(4-3i)}{4^2-(3i)^2} $$

$$ z=\frac{(3+2i)(1-2i)(4-3i)}{16+9} $$

$$ z=\frac{(3+2i)(1-2i)(4-3i)}{25} $$

Now, we can find the conjugate and magnitude of $$ z $$.

The conjugate of $$ z $$ is $$ \overline{z} $$, which is obtained by changing the sign of the imaginary part.

The magnitude of $$ z $$ is $$ |z| $$, which is the square root of the sum of the squares of the real and imaginary parts.

Let's calculate the conjugate and magnitude of $$ z $$ using the simplified expression.
Simplify the expression by following steps:
- step0: Solution:
  $$\frac{\frac{\left(3+2i\right)\left(1-2i\right)\left(4-3i\right)}{\left(4+3i\right)}}{\left(4-3i\right)}$$
- step1: Remove the parentheses:
  $$\frac{\frac{\left(3+2i\right)\left(1-2i\right)\left(4-3i\right)}{4+3i}}{4-3i}$$
- step2: Multiply the terms:
  $$\frac{\frac{16-37i}{4+3i}}{4-3i}$$
- step3: Divide the terms:
  $$\frac{-\frac{47}{25}-\frac{196}{25}i}{4-3i}$$
- step4: Multiply by the Conjugate:
  $$\frac{\left(-\frac{47}{25}-\frac{196}{25}i\right)\left(4+3i\right)}{\left(4-3i\right)\left(4+3i\right)}$$
- step5: Calculate:
  $$\frac{16-37i}{\left(4-3i\right)\left(4+3i\right)}$$
- step6: Calculate:
  $$\frac{16-37i}{25}$$
- step7: Simplify:
  $$\frac{16}{25}-\frac{37}{25}i$$
The simplified expression for $$ z $$ is $$ \frac{16}{25} - \frac{37}{25}i $$.

Now, we can find the conjugate and magnitude of $$ z $$.

The conjugate of $$ z $$ is $$ \overline{z} = \frac{16}{25} + \frac{37}{25}i $$.

The magnitude of $$ z $$ is $$ |z| = \sqrt{\left(\frac{16}{25}\right)^2 + \left(\frac{37}{25}\right)^2} $$.

Let's calculate the magnitude of $$ z $$.
Calculate the value by following steps:
- step0: Calculate:
  $$\sqrt{\left(\frac{16}{25}\right)^{2}+\left(\frac{37}{25}\right)^{2}}$$
- step1: Add the numbers:
  $$\sqrt{\frac{13}{5}}$$
- step2: Use the properties of radicals:
  $$\frac{\sqrt{13}}{\sqrt{5}}$$
- step3: Multiply by the Conjugate:
  $$\frac{\sqrt{13}\times \sqrt{5}}{\sqrt{5}\times \sqrt{5}}$$
- step4: Multiply the numbers:
  $$\frac{\sqrt{65}}{\sqrt{5}\times \sqrt{5}}$$
- step5: Multiply the numbers:
  $$\frac{\sqrt{65}}{5}$$
The magnitude of $$ z $$ is $$ \frac{\sqrt{65}}{5} $$.

Therefore, the conjugate of $$ z $$ is $$ \frac{16}{25} + \frac{37}{25}i $$ and the magnitude of $$ z $$ is $$ \frac{\sqrt{65}}{5} $$.

(b) If $$ z $$ is a complex number, then $$ |z| \geq 0 $$ and $$ |z| = 0 $$ if and only if $$ z = 0 $$.

This statement is true because the magnitude of a complex number is always non-negative, and it is zero only when the complex number is zero.
(a) The conjugate of $$ z $$ is $$ \frac{16}{25} + \frac{37}{25}i $$ and the magnitude of $$ z $$ is $$ \frac{\sqrt{65}}{5} $$. (b) If $$ z $$ is a complex number, then $$ |z| \geq 0 $$ and $$ |z| = 0 $$ if and only if $$ z = 0 $$.

1. (a) Find conjugate and magnitude of \( z=\frac{(3+2 i)(1-2 i)}{4+3 i} \) (To be solved in elass) (b) If \( = \) is a complex number then \( |z| \geq 0 \) and \( |z|=0 \) iff \( z=0 \) (To be solved in class) \( z=3(1-2 i)+2 i(1-2 i) \)

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