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If z1=2-i, z2=1+i, find abs((z1+z2+1)/(z...

If `z_1=2-i, z_2=1+i,` find `abs((z_1+z_2+1)/(z_1-z_2+i))`

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The correct Answer is:
`sqrt2`
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PATHFINDER-COMPLEX NUMBER-QUESTION BANK
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  10. If log(tan30^@)((2absz^2+2absz-3)/(absz+1))lt(-2), then

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  11. The number of solutions of the equation z^2+absz^2=0, where zinc is

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