ABS, ARG, CONJ, RECIPROCAL

Let z=-√2-√2i where i2 = -1. Find |z|, arg(z), z* (where * indicates the complex conjugate), and (1/z). Where appropriate, write your answers in the form a + i b,
where both a and b are real numbers.
Indicate the positions of z, z*, and (1/z) on an Argand diagram.

Final Answer:

a =  2
φ =  0.7854
Φ =  45 °
c = -√2+√2i
d = -√2/4 + √2/4 i

Step-by-step explanation:

$$\begin{gathered} z=-\surd 2-\surd 2i=Re+Im\cdot i=x+y\cdot i \\ x=-\sqrt{2}=-\sqrt{2}\doteq -1.4142 \\ y=-\sqrt{2}=-\sqrt{2}\doteq -1.4142 \\ a=\sqrt{R{e}^2+I{m}^2} \\ a=\sqrt{x^2+y^2}=\sqrt{(-1.4142{)}^2+(-1.4142{)}^2}=2 \end{gathered}$$

$$\begin{gathered} \tan \phi =y:x=(-1.4142):(-1.4142)=1=1:1 \\ \phi =\arctan (y/x)=\arctan ((-1.4142)/(-1.4142))=0.7854 \end{gathered}$$

$\Phi =\phi \to \text{°}=\phi \cdot \frac{180}{\pi }\text{°}=0.7854\cdot \frac{180}{\pi }\text{°}=45\text{°}=45^{\circ }$

$$\begin{gathered} c=conjz=x-y\cdot i \\ c=-\surd 2+\surd 2i \end{gathered}$$

$$\begin{gathered} d=1/z \\ d=\frac{1}{x+i\cdot y}=\frac{1}{x+i\cdot y}\cdot \frac{x-i\cdot y}{x-i\cdot y} \\ d=\frac{x-i\cdot y}{x^2-y^2} \\ X=\frac{x}{x^2+y^2}=\frac{(-1.4142)}{(-1.4142{)}^2+(-1.4142{)}^2}\doteq -0.3536 \\ Y=\frac{-y}{x^2+y^2}=\frac{-(-1.4142)}{(-1.4142{)}^2+(-1.4142{)}^2}\doteq 0.3536 \\ d=-\surd 2/4+\surd 2/4i \end{gathered}$$

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