Complex numbers: sketching {z in C | |3x - 1 + i| <= 2}, {z in C | I((1 - i)z) = 0}

[eminemsdictionary]

a) [imath]\{z \in \Complex\; \large{\vert}\normalsize{ \; |\,3z - 1 + i\,| \leq 2\}}[/imath]

b) [imath]\{z \in \Complex\; \large{\vert}\normalsize{\; \image\left((1 - i)z\right) = 0\}}[/imath]

How do you sketch something like that?

 

[stapel]

a) [imath]\{z \in \Complex\; \large{\vert}\normalsize{ \; |\,3z - 1 + i\,| \leq 2\}}[/imath]

b) [imath]\{z \in \Complex\; \large{\vert}\normalsize{\; \image\left((1 - i)z\right) = 0\}}[/imath]

How do you sketch something like that?

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Thank you!

Eliz.

 

[pka]

a) [imath]\{z \in \Complex\; \large{\vert}\normalsize{ \; |\,3z - 1 + i\,| \leq 2\}}[/imath]

b) [imath]\{z \in \Complex\; \large{\vert}\normalsize{\; \image\left((1 - i)z\right) = 0\}}[/imath]

Lets begin with some rather standard notation. If each of [imath]x~\&~y[/imath] is a real number
then [imath]z= x+yi[/imath] is a complex number and real part of [imath]z,~\Re(z)=x\text{ and the imagery part }\Im(z)=y[/imath].
The modulus or absolute value is [imath]|z|=\sqrt{\Re^2(z)+\Im^2(z})[/imath]. Because [imath]3z-1+i=(3x-1)+(3y+1)i[/imath]
we have [imath]|3z-1+i|=\sqrt{(3x-1)^2+(3y+1)^2}[/imath] That the necessary part for a).
For part b): [imath](1-i)(x+yi)=(x+y)+(y-x)i[/imath] thus [imath]\Im((1-i)z)=y-x[/imath]

[imath][/imath][imath][/imath][imath][/imath]