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Can someone please see why I am getting this error?

Posted on 2009-03-30
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Last Modified: 2016-02-10
I am new to Matlab and I am getting the following error;
Warning: Imaginary parts of complex X and/or Y arguments ignored
> In create_hyperbola at 13
Warning: Imaginary parts of complex X and/or Y arguments ignored
> In create_hyperbola at 15

what does this mean and why is this?

thanks in advance
% Program to plot the hyperbola
% y^2/a^2 - x^2/b^2 = 1
% The hyperbolae are open up/down, so that x is the independent variable
% for plotting. (Using the form x^2/a^2 - y^2/b^2 = 1 requires that y be
% the independent variable, which is awkward programming-wise.)
clear % all variables
figure(1), hold off % start a new figure
set(gca,'FontSize',14) % adjust fontsize
xmax = 30; ymax = 20;
x = linspace(-xmax,xmax,1001); % array of x values for plot (why 1001?)
a = 5; b = 3;
y=sqrt(((-x.^2)./(b^2)+1).*a^2); % corresponding y values
plot(x,y)
hold on % add to current plot
xlabel('x')
ylabel('y')
title(['Hyperbola $y^2/a^2 - x^2/b^2 = 1$; $a$ = ', num2str(a), ...
', $b$ = ', num2str(b),'; (WR 1/21/08)'],'Interpreter','latex')
% Add axes
plot([0 0],[-ymax ymax],'k') % y axis (black line - k)
plot([-xmax xmax],[0 0],'k') % x axis

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Question by:jtiernan2008
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by:yuk99
ID: 24021978
This is because when you calculating y (line 12) you getting square root mostly from negative numbers. So your y is vector of complex numbers. You get this because when you do -x.^2, it gets square first, than adds minus.
Do (-x).^2.
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Accepted Solution

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yuk99 earned 500 total points
ID: 24022077
Actually you will get different result. You probably just need what you get, but worry about the warning? Just ignore it, or you can get real part from complex y with real(y).

So in your code line 13 use:
plot(x,real(y))
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Author Closing Comment

by:jtiernan2008
ID: 31564470
thanks a million
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