+0

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

0
61
4
+355

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

waffles  Oct 28, 2017
Sort:

#1
+90999
+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?

Melody  Oct 28, 2017
#2
+355
+1

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

Guest Oct 29, 2017
#4
+18712
+1

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

where a and b are real numbers.

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

heureka  Oct 30, 2017

### 14 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details