1708\hspace{-16}$Here $\bf{\mathbb{Z}=(3+7i).(p+iq)=(3p-7q)+i.(7p+3q)}$\\\\\\ Now Here $\bf{\mathbb{Z}}$ is Purely Imiginary, Then $\bf{\mathbb{R\text{e}(Z)}=0}$\\\\\\ So $\bf{3p-7q=0\Leftrightarrow p=\frac{7}{3}.q}$ and $\bf{\mathbb{Z}=i.(7p+3q)}$\\\\\\ So $\bf{Min.\mid\mathbb{Z}^2 \mid = Min.\mid\mathbb{Z} \mid^2 = \mid 7p+3q \mid^2}$\\\\\\ Now Here $\bf{p\;,q\in \mathbb{Z}-\left\{0\right\}}$ and $\bf{p=\frac{7}{3}q}$\\\\\\ So We Take $\bf{p_{min}=\pm 3}$ and $\bf{q_{min}=\pm 7}$\\\\\\ Means $\bf{(p,q)=(3,7)}$ or $\bf{(p,q)=(-3,-7)}$\\\\\\ So $\bf{Min.\mid\mathbb{Z} \mid^2 = \mid 7p+3q \mid^2 = \mid 7.7+3.3\mid^2 = \mid -7.7-3.3\mid^2 = 3364}$
11z=(3+7i) (p+iq)=(3p-7q)+i(3q+7p)
now it is said z is purely imaginary
so,
3p-7q=0
or p/q=7/3
or p/q + i = 7/3 +i
(p+qi)/q=7+3i/3
from here the p+iq = 7+3i
z= 58i
|z2|= 3364
11epsilon,
p+iq = 7+3i means p=7 &q=3
how can you derive this from p/q=7/3 ??
i can take p=14&q=6 to get (p+qi)/q=7+3i/3
man111, plese explain your 2nd last & 3rd last line
11
epsilon
·2012-06-01 23:09:11
are yaar I hav not taken that way..
p/q + i = 7/3 +i
(p+qi)/q=7+3i/3
now
p+iq=7+3i
p=7
q=3
understood
:p
11
sougata nag
·2012-06-01 23:23:12
let me start from your 2nd line
(p+qi)/q=7+3i/3
now can't i say p=14 & q=6 ????
of course u r right in case of minimum.
11
epsilon
·2012-06-01 23:33:28
are yaar tum samjh te kyun nehi q is real and p+iq is imaginary.....
so...
imaginary no. only give imaginary results right thats why
p+iq = 7+3i
p=7
q=3
tum mujhe chat rahe ho??
11
sougata nag
·2012-06-01 23:44:23
p+iq = 7+3i implies that p=7 & q=3, i agree.
but (p+qi)/q=7+3i/3 does't imply that p=7 & q=3
IS IT CLEAR?
if i take your approach,then (p+qi)/q=7+3i/3=14+6i/6=21+9i/9=.....and so on
so i get{p=14,q=6},{p=21,q=9},....and so on.
not only that.
(p+qi)/q=7+3i/3=7x+3xi/3x
so p+qi=7x+3xi which implies that p=7x&q=3x for all real x.
NOW, CAN YOU DISAGREE OR U THINK KI MAI TUMHE CHAT RAHA HU?
