I try for this ... I try for this

I want to see the result like this

Plots from web

Plot[(-1.047-1.50133 I)*E^((-0.34381+1.35844 I) t) *(1+ Erf[(0.72714+0.9341 
I)* Sqrt(t))+((-1.047+1.50133 I)*E^((-0.34381-1.35844 I) t)*
1+Erf[(0.72714-0.9341 I)*Sqrt(t))]

closed as off-topic by Sjoerd C. de Vries, bbgodfrey, Öskå, Bob Hanlon, m_goldberg May 23 '15 at 18:55

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question arises due to a simple mistake such as a trivial syntax error, incorrect capitalization, spelling mistake, or other typographical error and is unlikely to help any future visitors, or else it is easily found in the documentation." – bbgodfrey, Öskå, Bob Hanlon
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 4
    $\begingroup$ Is it supposed that people willing to help will rewrite the formula from an image? $\endgroup$ – m0nhawk May 23 '15 at 12:17
  • $\begingroup$ possible duplicate of Plotting Complex Quantity Functions $\endgroup$ – m0nhawk May 23 '15 at 12:18
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  • 1
    $\begingroup$ As noted by @m0nhawk, please enter your code in the question with Mathematica format. $\endgroup$ – bbgodfrey May 23 '15 at 12:25
  • $\begingroup$ @m0nhawk yes,because the image is from the wolframalpha.com ,but I want to use Mathematica to show it. I also try the method of "Plotting Complex Quantity Functions " still can't plot it. $\endgroup$ – Erergo May 23 '15 at 12:40

In seems that your function has an imaginary part of zero for all inputs in range:

expr[t_] := (-1.047 - 1.50133 I)*
   E^((-0.34381 + 1.35844 I) t)*(1 + Erf[(0.72714 + 0.9341 I) *Sqrt[t]]) + (-1.047 + 
     1.50133 I)*E^(( -0.34381 - 1.35844 I) t)*(1 + Erf[(0.72714 - 0.9341  I)*Sqrt[t]])

Table[Im[expr[t]], {t, 0, 15, 0.1}] // Union

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