crystalbrite
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Im having trouble simplifying this complex number sqrt( (-1+i)/(sqrt2) ) into the form of x+iy. Could someone help please.
Simplyfying Complex Number: sqrt( (-1+i)/(sqrt(2)) )
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Re: Simplyfying Complex Number
Write ?(1-i) - using CiS form - then simplify.
crystalbrite said:
Write ?(1-i) - using CiS form - then simplify.
Re: Simplyfying Complex Number
Using polor form would be easier (if you are familiar with trig and the needed identities for simplifying), but you could take the square root as follows:
\(\displaystyle \sqrt{-1+i} = a+bi \text{ where}\)
\(\displaystyle a^2-b^2 = -1 \text{ and } 2ab=1\).
We get
\(\displaystyle a = \sqrt{b^2-1} \text{ or } a = -\sqrt{b^2-1} \,\, \Rightarrow \,\, 4b^2(b^2-1) = 1\)
\(\displaystyle 4b^4 - 4b^2 - 1 = 0 \,\, \Rightarrow \,\, b^2 = \frac{1 \pm \sqrt{2}}{2} \,\, \Rightarrow \,\, b = \frac{\sqrt{1+\sqrt{2}}}{\sqrt{2}}\)
Note I excluded the minus part of the "plus or minus". b^2 is a positive real number and 1-sqrt(2) is negative.
Now find a:
\(\displaystyle a = \frac{1}{2b} = \frac{1}{2\frac{\sqrt{1+\sqrt{2}}}{\sqrt{2}}} = \frac{1}{\sqrt{2+2\sqrt{2}}}\)
So, \(\displaystyle \frac{a+bi}{\sqrt{2}} = \frac{a}{\sqrt{2}} + \frac{b}{\sqrt{2}}i\)
\(\displaystyle \frac{\sqrt{-1+i}}{\sqrt{2}} = \frac{1}{2\sqrt{1+\sqrt{2}}} + \frac{\sqrt{1+\sqrt{2}}}{2}i\)
Using polor form would be easier (if you are familiar with trig and the needed identities for simplifying), but you could take the square root as follows:
\(\displaystyle \sqrt{-1+i} = a+bi \text{ where}\)
\(\displaystyle a^2-b^2 = -1 \text{ and } 2ab=1\).
We get
\(\displaystyle a = \sqrt{b^2-1} \text{ or } a = -\sqrt{b^2-1} \,\, \Rightarrow \,\, 4b^2(b^2-1) = 1\)
\(\displaystyle 4b^4 - 4b^2 - 1 = 0 \,\, \Rightarrow \,\, b^2 = \frac{1 \pm \sqrt{2}}{2} \,\, \Rightarrow \,\, b = \frac{\sqrt{1+\sqrt{2}}}{\sqrt{2}}\)
Note I excluded the minus part of the "plus or minus". b^2 is a positive real number and 1-sqrt(2) is negative.
Now find a:
\(\displaystyle a = \frac{1}{2b} = \frac{1}{2\frac{\sqrt{1+\sqrt{2}}}{\sqrt{2}}} = \frac{1}{\sqrt{2+2\sqrt{2}}}\)
So, \(\displaystyle \frac{a+bi}{\sqrt{2}} = \frac{a}{\sqrt{2}} + \frac{b}{\sqrt{2}}i\)
\(\displaystyle \frac{\sqrt{-1+i}}{\sqrt{2}} = \frac{1}{2\sqrt{1+\sqrt{2}}} + \frac{\sqrt{1+\sqrt{2}}}{2}i\)