#sketching complex
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can you draw the line Im(z) - Re(z) = 1?
No …
that's y - x = 1 on a ordinary cartesian plane
I might be cooked
Oh
Would that solve my issue?
And then shading the area within those 3 lines
yep
Really quickly, could you please give me some pointers (if you have time) on how I could attempt part b)?
This is where I’m at right now
The vertical segment is rather straightforward, varying the imaginary part would making the graph of the exponential draw an arc centred at the origin
The horizontal segment is also similar, should be a radial segment which, when extended, passes through the origin
The toughest part is the remaining logarithmic spiral mapped from the upward sloping segment
How do you recognise how to transform the function in this way
Are you take the exponent of the line?
As z—>e^z
The function is the transform
To be precise, I am taking the exponential of each point on the line segments
Since the real parts of all points on a vertical line are the same, they are mapped to points of equal magnitude
Similarly, the imaginary parts of all points on a horizontal line are the same, so they are mapped to points of equal angular argument
I'm sorry but im not too sure I understand. So the Im(z) and Re(z) stay the same?
But the y-x=1 line is altered by the exponential?
How would I sketch the remaining logarithmic spiral? Would that stem from the y-x=1 line?
ℝafain
try it out for specific points on the line (-1, pi/2 - 1 + pi/2 i, i, etc.)
What do you mean by try it out?
Map those specific points