Hello, i have this problem , how would i go at simplifying this, can i just divide the whole equation by ​?

You must have used sqrt2i) instead of sqrt(2)i to get your result Guest #5.

ok answer is 1+i nvm

Simplify the following:
(1 - sqrt(2) + 1 - sqrt(2 i))/(1 - sqrt(2 i))

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

Cancel exponents. sqrt((1 + i)^2) = 1 + i:
(1 - sqrt(2) + 1 - 1 + i)/(1 - sqrt(2 i))

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

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

Cancel exponents. sqrt((1 + i)^2) = 1 + i:
(1 - i - sqrt(2))/(1 - 1 + i)

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

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

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

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

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

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

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

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

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

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

-(-1 - i) = i + 1:
Answer: |1 + i - i sqrt(2)

\frac{((1-\sqrt{2})+(1-\sqrt{2}i)}{1-\sqrt{2}i}

$\frac{((1-\sqrt{2})+(1-\sqrt{2}i)}{1-\sqrt{2}i}\\ =\frac{(2-\sqrt{2}-\sqrt{2}i)}{1-\sqrt{2}i}\\ =\frac{(2-\sqrt{2}-\sqrt{2}i)}{1-\sqrt{2}i}\times \frac{1+\sqrt2i}{1+\sqrt2i}\\ =\frac{(1+\sqrt2i)(2-\sqrt{2}-\sqrt{2}i)}{1^2-(\sqrt{2}i)^2}\\ =\frac{(2-\sqrt{2}-\sqrt{2}i)+(\sqrt2i)(2-\sqrt{2}-\sqrt{2}i)}{1-(2*-1)}\\ =\frac{(2-\sqrt{2}-\sqrt{2}i)+(2\sqrt2i-\sqrt{2}\sqrt2i-\sqrt{2}\sqrt2ii)} {3}\\ =\frac{2-\sqrt{2}-\sqrt{2}i+2\sqrt2i-2i+2} {3}\\ =\frac{4-\sqrt{2}+\sqrt2i-2i} {3}\\ =\frac{4-\sqrt{2}+(\sqrt2-2)i} {3}\\ $

Hello, i have this problem
$\frac{(1-\sqrt{2})+(1-\sqrt{2}i)}{1-\sqrt{2}i}$
, how would i go at simplifying this.

$\begin{array}{|rcll|} \hline && \frac{(1-\sqrt{2})+(1-\sqrt{2}i)}{1-\sqrt{2}i} \\ &=& \frac{1-\sqrt{2}} {1-\sqrt{2}i} + \frac{1-\sqrt{2}i} {1-\sqrt{2}i} \\ &=& \frac{1-\sqrt{2}} {1-\sqrt{2}i} + 1 \\ &=& \frac{(1-\sqrt{2})} {(1-\sqrt{2}i)}\cdot \frac{(1+\sqrt{2}i)}{(1+\sqrt{2}i)} + 1 \\ &=& \frac{1\cdot (1+\sqrt{2}i)-\sqrt{2}(1+\sqrt{2}i)} {(1-\sqrt{2}i)(1+\sqrt{2}i)} +1 \\ &=& \frac{ 1+\sqrt{2}i-\sqrt{2} -\sqrt{2}\sqrt{2}i} {(1-\sqrt{2}i)(1+\sqrt{2}i)} +1 \\ &=& \frac{ 1-\sqrt{2}+\sqrt{2}i -2i} {(1-\sqrt{2}i)(1+\sqrt{2}i)} +1 \\ &=& \frac{ 1-\sqrt{2}+(\sqrt{2}-2)i } {(1-\sqrt{2}i)(1+\sqrt{2}i)} +1 \\ &=& \frac{ 1-\sqrt{2}+(\sqrt{2}-2)i } {[1^2-(\sqrt{2}i)^2]} +1 \\ &=& \frac{ 1-\sqrt{2}+(\sqrt{2}-2)i } {1 - 2i^2} +1 \quad & | \quad i^2 = -1\\ &=& \frac{ 1-\sqrt{2}+(\sqrt{2}-2)i } {1 -2\cdot (-1)} +1 \\ &=& \frac{ 1-\sqrt{2}+(\sqrt{2}-2)i } {1+2} +1 \\ &=& \frac{ 1-\sqrt{2}+(\sqrt{2}-2)i } {3} +1 \\ &=& \frac{ 1-\sqrt{2}+(\sqrt{2}-2)i } {3} +\frac33 \\ &=& \frac{ 4-\sqrt{2}+(\sqrt{2}-2)i } {3} \\ &=& \frac{ 4-\sqrt{2}+(\sqrt{2}-2)i } {3} \\ &=& \frac{ 4-\sqrt{2}}{3}+\frac{(\sqrt{2}-2) } {3}i \\ &=& 0.861928812542301650399437... - 0.195262145875634983732770... i \\ \hline \end{array}$

Melody and heureka:

Why does Wolfram/Alpha give this, very different, numerical result from both your answers?

1 - 0.4142135623... i

Put it into Wolframalpha and then copy and paste the link into a new post on this thread. 

Then we'll be able to see what you have done and we will probaby see why the answer is different.  :)

If you were a member you could message us the link as older posts can sometimes be hard to find ://

But we will probably see it anyway.  :)

OK. Thanks Alan. I included 2i under the sqrt sign, instead of just 2. Sorry about that.