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Trying to plot complex function for varying value of t

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Keaton Looper
Keaton Looper 2024 年 3 月 17 日
コメント済み: Walter Roberson 2024 年 3 月 20 日

My function is given by exp(i*t) and I want to plot the function for varying values of t. The function should show that it curves closer and closer to the imaginary axis as t increase from 0 to pi how would I show this and graph each increment on a graph? I have put an image using Desmos to show what I want matlab to graph.

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Keaton Looper
Keaton Looper 2024 年 3 月 17 日
Yes I want to show how for each value of t from 0 to pi the function curve gets closer to the imaginary axis as seen in the image above
dpb
dpb 2024 年 3 月 17 日
You're going to have to show us how you did what you did in Desmos; it doesn't look like what you describe to me.
clear all
t=linspace(0,pi); % vary t from 0 to pi
plot(exp(i*t)) % plot the function against t
ylim(xlim) % put on even range
axis square % and the same footing in space

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回答 (1 件)

Binaya
Binaya 2024 年 3 月 20 日
Hi Keaton
By Euler's formula, use of the expression "exp(i*t)" results in the expression "cos(t)+i*sin(t)". From this expression, it can be concluded that the maximum value of the expression can be 1 in either real axis or imaginary axis.
When you try to plot this "exp(i*t)", where t ranges from 0 to pi, you get a half circle as shown below:
t=0:0.01:pi;
y=exp(i*t);
plot(y);axis equal;
The plot generated by desmos does not seem to take into account the "i" or imaginary number when calculating the plot of the exponential which leads to the a plot of exponential function with real inputs.
  1 件のコメント
Walter Roberson
Walter Roberson 2024 年 3 月 20 日
As a special case, if you plot() a single variable and the variable is complex-valued, then it plots imaginary component of the value versus real component of the value.

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