Question

pls tell step by step solution , will follow and rate 5 star to right ans​

Answer 1

Step-by-step explanation:

We have,

$z_{1} = r_{1}( \cos \theta_{1} + i \sin \theta_{1}) \\ z_{2} = r_{2}( \cos \theta_{2} + i \sin \theta_{2})$

Now,

$\frac{z_{1}}{ z_{2} } = \frac{r_{1}}{ r_{2} } \bigg( \frac{ \cos \theta_{1} + i \sin \theta_{1}}{ \cos \theta_{2} + i \sin \theta_{2} } \bigg)$

$\implies \frac{z_{1}}{ z_{2} } = \frac{r_{1}}{ r_{2} } \bigg \{ \frac{ (\cos \theta_{1} + i \sin \theta_{1})( \cos \theta_{2} - i \sin \theta_{2} )}{( \cos \theta_{2} + i \sin \theta_{2} )( \cos \theta_{2} - i \sin \theta_{2} )} \bigg \}$

$\implies \frac{z_{1}}{ z_{2} } = \frac{r_{1}}{ r_{2} } \bigg \{ \frac{ (\cos \theta_{1}\cos \theta_{2} + i \sin \theta_{1}\cos \theta_{2} - i \cos \theta_{1} \sin \theta_{2} + \sin \theta_{1}\sin \theta_{2})}{ \cos^{2} \theta_{2} + \sin ^{2} \theta_{2} } \bigg \}$

$\implies \frac{z_{1}}{ z_{2} } = \frac{r_{1}}{ r_{2} } (\cos \theta_{1}\cos \theta_{2} +\sin \theta_{1}\sin \theta_{2} + i \sin \theta_{1}\cos \theta_{2} - i \cos \theta_{1} \sin \theta_{2} )$

$\implies \frac{z_{1}}{ z_{2} } = \frac{r_{1}}{ r_{2} } (\cos \theta_{1}\cos \theta_{2} +\sin \theta_{1}\sin \theta_{2} + i \sin \theta_{1}\cos \theta_{2} - i \cos \theta_{1} \sin \theta_{2} )$

$\implies \frac{z_{1}}{ z_{2} } = \frac{r_{1}}{ r_{2} } \{\cos (\theta_{1} - \theta_{2}) + i \sin (\theta_{1} - \theta_{2} ) \}$

So,

$\arg \bigg(\frac{z_{1}}{ z_{2} } \bigg)= (\theta_{1} - \theta_{2})$

Now,

$\bigg| \frac{z_{1}}{ z_{2} } \bigg| = \frac{r_{1}}{ r_{2} } \sqrt{\cos ^{2} (\theta_{1} - \theta_{2}) + \sin ^{2} (\theta_{1} - \theta_{2} ) }$

$\implies\bigg| \frac{z_{1}}{ z_{2} } \bigg| = \frac{r_{1}}{ r_{2} }$

Answer 2

Step-by-step explanation:

Correct option is

C

r=1,θ=tan

−1

(

4

3

)

2+i

1+2i

=r(cosθ+isinθ)

⟹

(2+i)(2−i)

(1+2i)(2−i)

=r(cosθ+isinθ)

⟹

5

4+3i

=r(cosθ+isinθ) ...(1)

⟹

∣

∣

∣

∣

∣

5

4+3i

∣

∣

∣

∣

∣

=r

∴r=1

From eq. (1)

r(cosθ+isinθ)=

5

4+3i