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# Express 1/1+/1/1-1/1+i in the form a+bi where a and b are real numbers.

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+644

Express 1/1+/1/1-1/1+i in the form a+bi where a and b are real numbers.

Oct 28, 2017

#1
+94991
+1

Express 1/1+/1/1-1/1+i in the form a+bi where a and b are real numbers.

What is 1+/1 supposed to mean?

Oct 28, 2017
#2
+644
0

Sorry if that didnt make any sense, here try this 1/(1+1/(1-1/(1-i)))

waffles  Oct 29, 2017
#3
+1

Simplify the following:
1/(1/(1 - 1/(-i + 1)) + 1)

Multiply numerator and denominator of (-1)/(-i + 1) by 1 + i:
1/(1/((-(i + 1))/((-i + 1) (i + 1)) + 1) + 1)

(-i + 1) (i + 1) = 1×1 + 1 i - i×1 - i×i = 1 + i - i + 1 = 2:
1/(1/(1 - ((i + 1))/2) + 1)

Put each term in 1 - (i + 1)/2 over the common denominator 2: 1 - (i + 1)/2 = 2/2 - (i + 1)/2:
1/(1/(2/2 + (-i - 1)/2) + 1)

Factor -1 from -i - 1:
1/(1/(2/2 + (-(i + 1))/2) + 1)

2/2 - (i + 1)/2 = (2 + (-i - 1))/2:
1/(1/((2 - 1 - i)/2) + 1)

Multiply the numerator of 1/((2 - 1 - i)/2) by the reciprocal of the denominator. 1/((2 - 1 - i)/2) = (1×2)/(2 - 1 - i):
1/(2/(2 - 1 - i) + 1)

2 - 1 - i = (2 - 1) - i = 1 - i:
1/(2/(-i + 1) + 1)

Multiply numerator and denominator of 2/(-i + 1) by 1 + i:
1/((2 (i + 1))/((-i + 1) (i + 1)) + 1)

(-i + 1) (i + 1) = 1×1 + 1 i - i×1 - i×i = 1 + i - i + 1 = 2:
1/((2 (i + 1))/2 + 1)

(2 (i + 1))/2 = 2/2×(i + 1) = i + 1:
1/(i + 1 + 1)

1 + i + 1 = (1 + 1) + i = 2 + i:
1/(i + 2)

Multiply numerator and denominator of 1/(i + 2) by 2 - i:
(-i + 2)/((i + 2) (-i + 2))

(i + 2) (-i + 2) = 2×2 + 2 (-i) + i×2 + i (-i) = 4 - 2 i + 2 i + 1 = 5:
(-i + 2)/5

Oct 29, 2017
#4
+20805
+1

Express 1/(1+1/(1-1/(1-i))) in the form a+bi

where a and b are real numbers.

$$\begin{array}{|rcll|} \hline && \mathbf{\cfrac{1}{1+\cfrac{1}{1-\cfrac{1}{ 1-i }}}} \\\\ &=& \cfrac{1}{1+\cfrac{1}{\cfrac{1-i}{ 1-i }-\cfrac{1}{ 1-i }}} \\\\ &=& \cfrac{1}{1+\cfrac{1}{\cfrac{1-i-1}{ 1-i }}} \\\\ &=& \cfrac{1}{1+\cfrac{1-i}{ \not{1}-i-\not{1} }} \\\\ &=& \cfrac{1}{1+\cfrac{1-i}{ -i} } \\\\ &=& \cfrac{1}{\cfrac{-i}{-i}+\cfrac{1-i}{-i} } \\\\ &=& \cfrac{1}{\cfrac{-i+1-i}{-i} } \\\\ &=& \dfrac{-i}{-i+1-i} \\\\ &=& \dfrac{-i}{1-2i} \\\\ &=& \left(\dfrac{-i}{1-2i} \right) \cdot \left( \dfrac{1+2i}{1+2i} \right) \\\\ &=& \dfrac{-i(1+2i)}{(1-2i)(1+2i)} \\\\ &=& \dfrac{-i-2i^2}{1-4i^2} \quad & | \quad i^2 = -1 \\\\ &=& \dfrac{-i-2(-1)}{1-4(-1)} \\\\ &=& \dfrac{-i+2}{1+4} \\\\ &=& \dfrac{-i+2}{5} \\\\ & \mathbf{=}& \mathbf{ \dfrac{2}{5}-\dfrac{1}{5}i } \\ \hline \end{array}$$

Oct 30, 2017