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How to delete/cancel trailing zeros in complex and imaginary numbers?

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Joshua Ford
Joshua Ford 2021년 5월 17일
댓글: Rik 2021년 5월 17일
Whenever I use or create imaginary and complex numbers, they are always saved with trailing zeros, for example 1.22000000 + 2.150000000i . I want to mimic the behaviour and structure of fixed-point while still being in floating-point. I want 1.22 + 2.15i. How do I delete or cancel the trailing zeros? I have tried different ways and as soon as the imaginary unit is present, the zeros come back.
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Stephen Cobeldick
Stephen Cobeldick 2021년 5월 17일
"I want 1.22 + 2.15i."
Why?
What specific operations are you performing that cannot be achieved using normal binary floating point numbers?

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답변(2개)

Rik
Rik 2021년 5월 17일
The way data is stored and the way it is displayed is not necessarilly connected. You can influence the way data is presented with the format function. If you want more control, you need the fprintf or sprintf functions.
format shortG
1.22 + 2.15i
ans =
1.22 + 2.15i
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Rik
Rik 2021년 5월 17일
You already have the values very close to that. You can round to a specific number of decimals, but due to what James wrote you can't get close than this:
round(1.223 + 2.148i,2)
ans = 1.2200 + 2.1500i

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Paul Hoffrichter
Paul Hoffrichter 2021년 5월 17일
편집: Paul Hoffrichter 2021년 5월 17일
a = 1.22000000 + 2.150000000i
a = 1.2200 + 2.1500i
fprintf("%g + %gi\n", real(a), imag(a));
1.22 + 2.15i
fprintf("%.2f + %.2fi\n", real(a), imag(a));
1.22 + 2.15i
  댓글 수: 3
Paul Hoffrichter
Paul Hoffrichter 2021년 5월 17일
>> Does this store the numbers like this or does it only display the numbers as shown?
The storage has not changed with the choice of display.
Below shows two different variables set to a value with and without trailing zeros. But the two number numbers are represented internally exactly the same way.
a = 1.22 + 2.15i;
b = 1.22000000 + 2.150000000i;
a == b
ans = logical
1

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