+0  
 
0
365
2
avatar

Given that z=-√2-√2i, calculate z^8.

Guest Mar 18, 2015

Best Answer 

 #1
avatar+19994 
+10

Given that z=-√2-√2i, calculate z^8

$$\small{\text{
$z= -\sqrt2-\sqrt2\cdot i \qquad
\boxed{z= -\sqrt2\cdot (1+i)}
$}}\\\\
\small{\text{
$
z^8 = \left[-\sqrt2 \cdot(1+i) \right] ^8
$}}\\
\small{\text{
$
z^8 = (-\sqrt2)^8\cdot (1+i)^8
$
}} \\
\small{\text{
$
z^8 = (-1)^8\cdot 2^{\frac82}\cdot (1+i)^8
$
}} \\
\small{\text{
$ z^8 = 1\cdot 2^4\cdot (1+i)^8 $
}} \\
\small{\text{
$
\begin{array}{l|l|l}
z^8 = 2^4\cdot (1+i)^8 \quad & \quad (1+i)^2 =1+2\cdot i+ i^2 \quad & \quad \boxed{\ i^2=-1 \ } \\
& \quad (1+i)^2 = 1+2\cdot i -1 \\
& \quad (1+i)^2 = 2\cdot i \\
z^8 = 2^4 \cdot \left[(1+i)^2\right]^4 \\
z^8 = 2^4 \cdot \left[ 2\cdot i\right]^4 \\
z^8 = 2^4 \cdot 2^4 \cdot i^4 \\
z^8 = 2^8\cdot i^4 \quad & \quad i^4=i^2\cdot i^2\\
& \quad i^4=(-1)\cdot (-1)\\
& \quad i^4 = (-1)^2 \\
& \quad i^4 = 1 \\
\boxed{\ z^8 = 2^8 = 256 \ }
\end{array}
$
}} \\$$

heureka  Mar 18, 2015
 #1
avatar+19994 
+10
Best Answer

Given that z=-√2-√2i, calculate z^8

$$\small{\text{
$z= -\sqrt2-\sqrt2\cdot i \qquad
\boxed{z= -\sqrt2\cdot (1+i)}
$}}\\\\
\small{\text{
$
z^8 = \left[-\sqrt2 \cdot(1+i) \right] ^8
$}}\\
\small{\text{
$
z^8 = (-\sqrt2)^8\cdot (1+i)^8
$
}} \\
\small{\text{
$
z^8 = (-1)^8\cdot 2^{\frac82}\cdot (1+i)^8
$
}} \\
\small{\text{
$ z^8 = 1\cdot 2^4\cdot (1+i)^8 $
}} \\
\small{\text{
$
\begin{array}{l|l|l}
z^8 = 2^4\cdot (1+i)^8 \quad & \quad (1+i)^2 =1+2\cdot i+ i^2 \quad & \quad \boxed{\ i^2=-1 \ } \\
& \quad (1+i)^2 = 1+2\cdot i -1 \\
& \quad (1+i)^2 = 2\cdot i \\
z^8 = 2^4 \cdot \left[(1+i)^2\right]^4 \\
z^8 = 2^4 \cdot \left[ 2\cdot i\right]^4 \\
z^8 = 2^4 \cdot 2^4 \cdot i^4 \\
z^8 = 2^8\cdot i^4 \quad & \quad i^4=i^2\cdot i^2\\
& \quad i^4=(-1)\cdot (-1)\\
& \quad i^4 = (-1)^2 \\
& \quad i^4 = 1 \\
\boxed{\ z^8 = 2^8 = 256 \ }
\end{array}
$
}} \\$$

heureka  Mar 18, 2015
 #2
avatar
+5

A general rule when working with complex numbers is , for addition and subtraction use the algebraic form of the number, for multiplication powers and roots use the polar or exponential form.

So, first putting the number into polar form,

$$-\sqrt{2}-\imath\sqrt{2}=-2(1/\sqrt{2}+\imath/\sqrt{2})=-2\angle45\deg$$

and now raising to the power eight using De Moivre's theorem,

$$(-2\angle45\deg)^{8}=(-2)^{8}\angle(8\times45\deg)=256\angle360\deg=256.$$

Guest Mar 18, 2015

34 Online Users

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.