A pure imaginary number is of the form \(bi\) where \(b\) is a [non-zero] real number and \(i^2=-1\).

If \(a\) real number were equal to \(bi\), then we would have \(a^2=-b^2\) which is impossible since the right hand side is negative while the left hand side is non-negative.
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Mursalin Habib
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2 years, 1 month ago

@Mursalin Habib
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No, you realy are awsome, also your set ''proof problem of the day'' is realy cool one
–
Aman Sharma
·
2 years, 1 month ago

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Actually, by the definition of complex numbers, a real number is also considered a complex number. Think of it as like, we can write x as x + 0i.
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Zi Song Yeoh
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2 years, 1 month ago

Sorry, all real numbers are complex numbers. If this is from NCERT, exercise and problem number please?
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Agnishom Chattopadhyay
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2 years, 1 month ago

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TopNewestA pure imaginary number is of the form \(bi\) where \(b\) is a [non-zero] real number and \(i^2=-1\).

If \(a\) real number were equal to \(bi\), then we would have \(a^2=-b^2\) which is impossible since the right hand side is negative while the left hand side is non-negative. – Mursalin Habib · 2 years, 1 month ago

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– Aman Sharma · 2 years, 1 month ago

Thanks for replying you are awsome in proof problemsLog in to reply

– Mursalin Habib · 2 years, 1 month ago

Thanks! But I wouldn't say I'm awesome.Log in to reply

– Aman Sharma · 2 years, 1 month ago

No, you realy are awsome, also your set ''proof problem of the day'' is realy cool oneLog in to reply

Actually, by the definition of complex numbers, a real number is also considered a complex number. Think of it as like, we can write x as x + 0i. – Zi Song Yeoh · 2 years, 1 month ago

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– Aman Sharma · 2 years, 1 month ago

I am editing it thanks for pointing out the mistakeLog in to reply

Sorry, all real numbers are complex numbers. If this is from NCERT, exercise and problem number please? – Agnishom Chattopadhyay · 2 years, 1 month ago

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– Aman Sharma · 2 years, 1 month ago

EditedLog in to reply