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Question Number 158378 by Rasheed.Sindhi last updated on 03/Nov/21

How to graph order pair (3+5i , 4−2i)?

$${How}\:{to}\:{graph}\:{order}\:{pair}\:\left(\mathrm{3}+\mathrm{5}{i}\:,\:\mathrm{4}−\mathrm{2}{i}\right)? \\ $$

Answered by MJS_new last updated on 03/Nov/21

you would need a 4−dimensional coordinate  system...

$$\mathrm{you}\:\mathrm{would}\:\mathrm{need}\:\mathrm{a}\:\mathrm{4}−\mathrm{dimensional}\:\mathrm{coordinate} \\ $$$$\mathrm{system}... \\ $$

Commented by ajfour last updated on 03/Nov/21

If f(z=x+iy)=p(x,y)+iq(x,y)  just if  p>0; q>0  always  we can have two surfaces  one for p green one, and other  q blue one.

$${If}\:{f}\left({z}={x}+{iy}\right)={p}\left({x},{y}\right)+{iq}\left({x},{y}\right) \\ $$$${just}\:{if}\:\:{p}>\mathrm{0};\:{q}>\mathrm{0}\:\:{always} \\ $$$${we}\:{can}\:{have}\:{two}\:{surfaces} \\ $$$${one}\:{for}\:{p}\:{green}\:{one},\:{and}\:{other} \\ $$$${q}\:{blue}\:{one}. \\ $$

Commented by Rasheed.Sindhi last updated on 03/Nov/21

Thanks to both (M/A)J(4/S) sirs!  Once I had tried to graph the   ordered pair of complex numbers  using two intersecting planes that  is in 3-dimentional space but I was  not completely successful...

$$\mathbb{T}\mathrm{han}\Bbbk\mathrm{s}\:\mathrm{to}\:\mathrm{both}\:\frac{\mathbb{M}}{\mathbb{A}}\mathbb{J}\frac{\mathrm{4}}{\mathbb{S}}\:\mathrm{sirs}! \\ $$$$\mathrm{Once}\:\mathrm{I}\:\mathrm{had}\:\mathrm{tried}\:\mathrm{to}\:\mathrm{graph}\:\mathrm{the}\: \\ $$$$\mathrm{ordered}\:\mathrm{pair}\:\mathrm{of}\:\mathrm{complex}\:\mathrm{numbers} \\ $$$$\mathrm{using}\:\mathrm{two}\:\mathrm{intersecting}\:\mathrm{planes}\:\mathrm{that} \\ $$$$\mathrm{is}\:\mathrm{in}\:\mathrm{3}-\mathrm{dimentional}\:\mathrm{space}\:\mathrm{but}\:\mathrm{I}\:\mathrm{was} \\ $$$$\mathrm{not}\:\mathrm{completely}\:\mathrm{successful}... \\ $$

Commented by MJS_new last updated on 03/Nov/21

we draw a cube in 2d by drawing 2 squares  and linking them with 4 bevel lines  I once saw the 3d model of a 4d hyper cube,  it consisted of 2 3d−cubes linked by 8 bevel  lines... if you try to draw this in 2d you lose  too much information, it′s impossible to see  what you wanted to show

$$\mathrm{we}\:\mathrm{draw}\:\mathrm{a}\:\mathrm{cube}\:\mathrm{in}\:\mathrm{2d}\:\mathrm{by}\:\mathrm{drawing}\:\mathrm{2}\:\mathrm{squares} \\ $$$$\mathrm{and}\:\mathrm{linking}\:\mathrm{them}\:\mathrm{with}\:\mathrm{4}\:\mathrm{bevel}\:\mathrm{lines} \\ $$$$\mathrm{I}\:\mathrm{once}\:\mathrm{saw}\:\mathrm{the}\:\mathrm{3d}\:\mathrm{model}\:\mathrm{of}\:\mathrm{a}\:\mathrm{4d}\:\mathrm{hyper}\:\mathrm{cube}, \\ $$$$\mathrm{it}\:\mathrm{consisted}\:\mathrm{of}\:\mathrm{2}\:\mathrm{3d}−\mathrm{cubes}\:\mathrm{linked}\:\mathrm{by}\:\mathrm{8}\:\mathrm{bevel} \\ $$$$\mathrm{lines}...\:\mathrm{if}\:\mathrm{you}\:\mathrm{try}\:\mathrm{to}\:\mathrm{draw}\:\mathrm{this}\:\mathrm{in}\:\mathrm{2d}\:\mathrm{you}\:\mathrm{lose} \\ $$$$\mathrm{too}\:\mathrm{much}\:\mathrm{information},\:\mathrm{it}'\mathrm{s}\:\mathrm{impossible}\:\mathrm{to}\:\mathrm{see} \\ $$$$\mathrm{what}\:\mathrm{you}\:\mathrm{wanted}\:\mathrm{to}\:\mathrm{show} \\ $$

Commented by Rasheed.Sindhi last updated on 03/Nov/21

You′re very right sir! I also experienced  similarly.I achieved graphs represent  either real parts or imaginary parts  but not both!   Thanks again sir!

$$\mathrm{You}'\mathrm{re}\:\mathrm{very}\:\mathrm{right}\:\mathrm{sir}!\:\mathrm{I}\:\mathrm{also}\:\mathrm{experienced} \\ $$$$\mathrm{similarly}.\mathrm{I}\:\mathrm{achieved}\:\mathrm{graphs}\:\mathrm{represent} \\ $$$$\mathrm{either}\:\mathrm{real}\:\mathrm{parts}\:\mathrm{or}\:\mathrm{imaginary}\:\mathrm{parts} \\ $$$$\mathrm{but}\:\mathrm{not}\:\mathrm{both}!\: \\ $$$$\mathbb{T}\mathrm{hanks}\:\mathrm{again}\:\mathrm{sir}! \\ $$

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