Question

if Re(z) = 0 then is z a purely real number or imaginary number

Answer 1

i hope its Imaginary number

Answer 2

yes u r right but how diya ?

Answer 3

Re(z) = Real part of z
let z be a+ib

Re(z)= a
im(z)=b

Re(z)=0 therefore a=0

z=0+ib =ib
and its an imaginary number

Answer 4

Re(z) = Real part of z
Im(z)=Imaginary part of z

Answer 5

did you understand?

Answer 6

yes diya thanks a lot

Answer 7

welcome:)

Answer 8

You forgot something, what if z=0? You have to separate in two cases.

Answer 9

Well ,isn't it given Re(z)=0 ?

Answer 10

When Re9Z0=0 the result is imaginary, on the other side if Im9z)=0, it is real number.Got it ?

Answer 11

1348 is right.
In the case Im(z)=0 then z=0+i0=0, which is in R.

Answer 12

SRY Re(z)

Answer 13

Though the implication is correct, since z is either real or complex, exclusively.

Answer 14

if Im(z) = 0, then z is a real number. On the other hand, if RE(z)=0 it is an imaginary number.

Answer 15

Two cases:
\[1. z \in \mathbb{C}^*\]
\[Re(z)=0 \iff z=re^{i \theta}, \theta \in ]-\pi ; \pi [ \setminus \left\{ 0 \right\}, r \in \mathbb{R}^* \\\]
\[\iff z \in i\mathbb{R} \]

BUT, case two...
\[z=0 \in \mathbb{R} \implies Re(z)=0\]
So no, your implication isn't direct. I can have a real number z with Re(z)=0, that number's called 0!

Answer 16

\[z \in i\mathbb{R} \implies Re(z) =0 \]
is true
BUT, here's the difference...
\[Re(z)=0 \implies z \in i\mathbb{R} \]
is false, because 0 is real and Re(0)=0

Answer 17

its imaginary number ,,but overall it is a complex number ,, as u know every number real or imaginary can be expressed in complex form so it is always suitable to say such numbers complex numbers ,,,but if u want to be specific it is imaginary number .

Answer 18

why can i give only 1 medal for u all thanks a lot

Answer 19

OS Changed :)