If z and are two non-zero complex numbers such that |z|=1 and Arg(z)-Arg()=2, then z is eq…

Question

If $z$ and $\omega$ are two non-zero complex numbers such that $|z\omega|=1$ and $\operatorname{Arg}(z)-\operatorname{Arg}(\omega)=\dfrac\pi2$, then $z\omega$ is equal to

Accepted Answer

Given that,

$$\left| {z\omega } \right| = 1$$

$$ \Rightarrow $$ $$\left| z \right|\left| \omega  \right|$$ = 1

$$ \Rightarrow $$ $$\left| z \right|$$ = $${1 \over {\left| \omega  \right|}}$$

and $$Arg(z) - Arg(\omega ) = {\pi  \over 2}$$

$$ \Rightarrow $$ $$Arg\left( {{z \over \omega }} \right)$$ $$= {\pi  \over 2}$$

When argument of a complex number is $${\pi  \over 2}$$, it means it is making an angle of $${\pi  \over 2}$$ with the real axis in the counterclockwise, so it is along the imaginary axis and positive side of imaginary axis.

So, $${{z \over \omega }}$$ is a purely imaginary number that means there is no real part in this complex number.

So we can assume,

$${{z \over \omega }}$$ = $ki$

$$ \Rightarrow $$ $${\left| {{z \over \omega }} \right|}$$ = $$\left| {ki} \right|$$

$$ \Rightarrow $$ $${\left| {{z \over \omega }} \right|}$$ = $$\left| k \right|\left| i \right|$$

$$ \Rightarrow $$ $${\left| {{z \over \omega }} \right|}$$ = $k$        [ as $$\left| i \right|$$ = 1 ]

$$ \Rightarrow $$ $$\left| z \right|$$ $ \times $$ ${1 \over {\left| \omega  \right|}}$$ = $k$

$$ \Rightarrow $$ $$\left| z \right|$$ $ \times $$\left| z \right|$=$k$$       [ as $${1 \over {\left| \omega  \right|}}$=$\left| z \right|$$ ]

$$ \Rightarrow $${\left| z \right|^2}$=$k$$ \Rightarrow $$\left| z \right|$=$\sqrt k $$\therefore$$\left| \omega  \right|$=${1 \over {\sqrt k }}$As${{z \over \omega }}$is imaginary so we can write,${{z \over \omega }}$=$ - {{\overline z } \over {\overline \omega  }}$$

[ When $$z$is imaginary then$z$=$-\overline z $$ ]

$$ \Rightarrow $$\overline z \omega $=$ - z\overline \omega  $$ \Rightarrow $$\overline z \omega $=$-{{z \over \omega }}$.$\overline \omega  $.$\omega $$ \Rightarrow $$\overline z \omega $=$-{{z \over \omega }}$.${\left| \omega  \right|^2}$$ \Rightarrow $$\overline z \omega $=$-ki$.${\left( {{1 \over {\sqrt k }}} \right)^2}$$ \Rightarrow $$\overline z \omega $=$-ki$.${1 \over k}$$ \Rightarrow $$\overline z \omega $=$-i$Method 2 :

Given that,$\left| {z\omega } \right| = 1$$ \Rightarrow $$\left| z \right|\left| \omega  \right|$= 1$ \Rightarrow $$\left| z \right|$=${1 \over {\left| \omega  \right|}}$and$Arg(z) - Arg(\omega ) = {\pi  \over 2}$$ \Rightarrow $$Arg\left( {{z \over \omega }} \right) ( \]= {\pi  \over 2}$When argument of a complex number is${\pi  \over 2}$, it means it is making an angle of${\pi  \over 2}$$ with the real axis in the counterclockwise, so it is along the imaginary axis and positive side of imaginary axis.

So, $${{z \over \omega }}$$ is a purely imaginary number that means there is no real part in this complex number.

So we can assume,

$${{z \over \omega }}$=$ki$$ \Rightarrow $${\left| {{z \over \omega }} \right|}$=$\left| {ki} \right|$$ \Rightarrow $${\left| {{z \over \omega }} \right|}$=$\left| k \right|\left| i \right|$$ \Rightarrow $${\left| {{z \over \omega }} \right|}$=$k$$        [ as $$\left| i \right|$$ = 1 ]

$$ \Rightarrow $$\left| z \right|$$ $ \times $$ ${1 \over {\left| \omega  \right|}}$=$k$$ \Rightarrow $$\left| z \right|$$ $ \times $$ $$\left| z \right|$$ = $k$       [ as $${1 \over {\left| \omega  \right|}}$$ = $$\left| z \right|$$ ]

$$ \Rightarrow $$ $${\left| z \right|^2}$$ = $k$

$$ \Rightarrow $$ $$\left| z \right|$$ = $$\sqrt k $$

$$\therefore$$ $$\left| \omega  \right|$$ = $${1 \over {\sqrt k }}$$

(1) Magnitude of $$\overline z \omega $$

= $$\left| {\overline z } \right|\left| \omega  \right|$$

= $$\left| z \right|\left| \omega  \right|$$ [ as $$\left| z \right|$$ = $$\left| {\overline z } \right|$$ ]

= $$\sqrt k $$.$${{1 \over {\sqrt k }}}$$

= 1

$$\therefore$$ The distance from the origin of $${\overline z \omega }$$ is 1.

(2) Argument of $${\overline z \omega }$$ = $$Arg\left( {\overline z \omega } \right)$$

= $$Arg\left( {\overline z } \right) + Arg\left( \omega  \right)$$

= $$-Arg\left( z \right) + Arg\left( \omega  \right)$$

= $$ - \left( {Arg\left( z \right) - Arg\left( \omega  \right)} \right)$$

= $$ - {\pi  \over 2}$$

$$\therefore$$ $${\overline z \omega }$$ is at (0, -1) on the negative side of imaginary axis and making an angle of $${\pi  \over 2}$$ clockwise.

$$\therefore$$ $${\overline z \omega }$$ = 0 + (-1) ( \times \] $i$=$-i$

Solutions · 1