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# Calculate z^8.

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Given that z=-√2-√2i, calculate z^8.

Guest Mar 18, 2015

#1
+19624
+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
+19624
+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
#2
+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