# Thread: Complex numbers?? does it go here?

1. ## Complex numbers?? does it go here?

solve the equations to find z and w, expressed in a + ib form

z + (1 - i)w = 2i
w + (1 - i)z = 1

Thanks

2. Originally Posted by differentiate
solve the equations to find z and w, expressed in a + ib form

z + (1 - i)w = 2i
w + (1 - i)z = 1

Thanks
Are z and w meant to represent unknown real numbers? (If so, this is very poor notation since both pronumerals are typically used to represent complex numbers).

3. Originally Posted by differentiate
solve the equations to find z and w, expressed in a + ib form

z + (1 - i)w = 2i
w + (1 - i)z = 1

Thanks
hello differentiate,
these are the system of linear equation in two variable w and z(i agree with mr.fantastic that this is wrong representation with complex numbers to use 'z' and 'w' as numeral variables). try to solve them by any method which is suitable to you like substitution method,Elimination method etc.and put i^2=(-1) where require.
if you have any confusion, look at spoiler;i have used Elimination method in spoiler.

Spoiler:
Code:


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$(1 - i)w+z = 2i \quad ............(1)$

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$w + (1 - i)z = 1 \quad ............(2)$
multiply (2) by (1 - i) and subtract from (1), we get

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$\{ (1 - i)w+z\} -\{ (1 - i)w + (1 - i)^2z \}=2i-(1-i)$

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$\Rightarrow \quad z-(1 - i)^2z=(3i-1)\quad \Rightarrow \quad z-(-2 i)z=(-1+3i)$

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$\Rightarrow \quad(1+2i)z=-1+3i\quad \Rightarrow \quad z=\frac{-1+3i}{1+2i}$
rationalize

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$z=\frac{-1+3i}{1+2i}$ we get

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$\boxed{z=1+i}$
substitute value of z in (2),we get

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$\boxed{w=-1}$

4. Thanks champions!
I get it now