Complex Numbers Ques 62
- $|z| \leq 1$, $|w| \leq 1$, then show that
$|z-w|^{2} =(|z|-|w|)^{2}+(\arg z-\arg w)^{2}$
(1995, 5M)
Show Answer
Solution:
Formula:
- Let $z=r _1\left(\cos \theta _1+i \sin \theta _1\right)$ and $w=r _2\left(\cos \theta _2+i \sin \theta _2\right)$ We have, $|z|=r _1,|w|=r _2, \arg (z)=\theta _1$ and $\arg (w)=\theta _2$ Given, $|z| \leq 1,|w|<1$
$\Rightarrow \quad r _1 \leq 1$ and $r _2 \leq 1$
Now,
$$ \begin{gathered} z-w=\left(r _1 \cos \theta _1-r _2 \cos \theta _2\right)+i\left(r _1 \sin \theta _1-r _2 \sin \theta _2\right) \\ \Rightarrow|z-w|^{2}=\left(r _1 \cos \theta _1-r _2 \cos \theta _2\right)^{2}+\left(r _1 \sin \theta _1-r _2 \sin \theta _2\right)^{2} \\ =r _1^{2} \cos ^{2} \theta _1+r _2^{2} \cos ^{2} \theta _2-2 r _1 r _2 \cos \theta _1 \cos \theta _2 \\ \quad+r _1^{2} \sin ^{2} \theta _1+r _2^{2} \sin ^{2} \theta _2-2 r _1 r _2 \sin \theta _1 \sin \theta _2 \\ =r _1^{2}\left(\cos ^{2} \theta _1+\sin ^{2} \theta _1\right)+r _2^{2}\left(\cos ^{2} \theta _2+\sin ^{2} \theta _2\right) \\ \quad-2 r _1 r _2\left(\cos \theta _1 \cos \theta _2+\sin \theta _1 \sin \theta _2\right) \\ =r _1^{2}+r _2^{2}-2 r _1 r _2 \cos \left(\theta _1-\theta _2\right) \\ =\left(r _1-r _2\right)^{2}+2 r _1 r _2\left[1-\cos \left(\theta _1-\theta _2\right)\right] \\ =\left(r _1-r _2\right)^{2}+4 r _1 r _2 \sin ^{2} \frac{\theta _1-\theta _2}{2} \\ \left.\leq\left|r _1-r _2\right|^{2}+4 \sin \frac{\theta _1-\theta _2}{2}, r _1 \leq 1\right] \end{gathered} $$
and $|\sin \theta| \leq|\theta|, \forall \theta \in R$
Therefore, $|z-w|^{2} \leq\left|r _1-r _2\right|^{2}+2 (\theta _1-\theta _2)$
$$ \leq\left|r _1-r _2\right|^{2}+\left|\theta _1-\theta _2\right|^{2} $$
$$ \Rightarrow \quad|z-w|^{2} \leq(|z|-|w|)^{2}+(\arg z-\arg w)^{2} $$
Alternate Solution
$$ \begin{gathered} |z-w|^{2}=|z|^{2}+|w|^{2}-2|z||w| \cos (\arg z+\arg w) \\ |z|^{2}+|w|^{2} -2|z||w| \cos (\arg z-\arg w) \\ =(|z|-|w|)^{2}+2|z||w| \cdot 2 \sin ^{2} \frac{\arg z-\arg w}{2} \quad \ldots(i) \\ \therefore \quad|z-w|^{2} \leq(|z|-|w|)^{2}+4 \cdot 1 \cdot 1 \frac{\arg z-\arg w}{2} \\ \Rightarrow \quad|z-w|^{2} \leq(|z|-|w|)^{2}+(\arg z-\arg w)^{2} \end{gathered} $$
BETA
