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Esay Complex Number Question (1 Viewer)

CLD

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Hi all,

How do you sketch Im(z) = |z|?

I think it is a circle?
 

nightweaver066

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When in doubt, let z = x + iy

Let z = x + iy











However |x^2 + y^2| 0, so the solution is the positive imaginary axis, i.e.
 
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jnney

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y = sqrt(x^2+y^2)

y^2=x^2+y^2

x^2=0

x=0
 

SpiralFlex

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"Beware the twraps of modulus" quoted from my side-kick.
 

jnney

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Okay but how can you sketch x=0 and y>(equal to)0 on one graph paper?
it's just a single line on the y=axis. Except:

y>=0 is a condition which the graph x=0 must satisfy. it means that this particular graph does not extend below the origin.
 

Carrotsticks

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Okay but how can you sketch x=0 and y>(equal to)0 on one graph paper?
x=0 implies the entire Y axis.

y greater than or equal to zero implies the top half of the xy plane (ie: Quadrants 1 and 2).

The only region satisfying both these conditions is the top half of the Y axis (including the origin).
 

school4nerds

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let z= x+iy,

Im(z)= y

|z|= (sqrt)(x^2 + y^2)

y^2= x^2+y^2

x^2=0

x=0

why is this wrong? i thought you could square both sides? :\
 

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