Question

Find the modulus and amplitude of complex number. 8+15i and express it into pollar form and exponential form
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Answer 1

Given expression,

$\sf \: z = 8 + 15 \iota$

Now,

$\sf \: lf \: z = a + b\iota \: then \: |z| = \sqrt{ {a}^{2} + {b}^{2} }$

Therefore,

$\sf \: |z| = \sqrt{8 {}^{2} + {15}^{2} } \\ \\ \longrightarrow \sf \: |z| = \sqrt{64 + 225} \\ \\ \longrightarrow \sf \: |z| = \sqrt{289} \\ \\ \longrightarrow \boxed{ \boxed{ \sf \: r = 17}}$

Amplitude will be :

Since, 8 and 15 lie in first quadrant.

$\sf \: \alpha = tan {}^{ - 1} ( \dfrac{15}{8} )$

Therefore,

$\sf \: tan \: \alpha = \dfrac{15}{8} \\ \\ \longrightarrow \sf \: \dfrac{sin \: \alpha }{cos \: \alpha } = \dfrac{ \dfrac{15}{ \sqrt{ {15}^{2} + {8}^{2} } } }{ \dfrac{8}{ \sqrt{ {15}^{2} + {8}^{2} } } } \\ \\ \longrightarrow \sf \: sin \: \alpha = \dfrac{15}{17} \: and \: cos \: \alpha = \dfrac{8}{17}$

Polar form :

$\sf \: z = r(cos \: \alpha + \iota \: sin\: \alpha ) \\ \\ \longrightarrow \boxed{ \boxed{ \sf \: z = 17(cos \: \alpha + \iota \: sin\: \alpha )}}$

Exponential Form :

$\sf \: z = re {}^{ \iota \alpha } \\ \\ \longrightarrow \boxed{\boxed{\sf z = 17e^{\iota \alpha}}}$

Answer 2

Answer:

$\implies{\boxed{\sf{r=17}}}$

$\implies{\boxed{\sf{Amplitude=tan^{-1}\Bigg(\dfrac{15}{8}\Bigg)}}}$

$\implies{\boxed{\sf{z\;(Polar\;Form)=17(\cos \theta + i\sin \theta)}}}$

$\implies{\boxed{\sf{z\;(Exponential\;form)=17\exp^{i\theta}}}}$

Step-by-step explanation:

Given:

• Complex number = 8 + 15i

To Find:

• Modulus
• Amplitude

Formula used:

• $\sf{|z|=\sqrt{a^{2}+b^{2}}}$

Let, z = 8 + 15i

Here, a = 8 and b = 15.

$\implies{\sf{|z|=r=\sqrt{a^{2}+b^{2}}}}$

$\implies{\sf{r=\sqrt{8^{2}+15^{2}}}}$

$\implies{\sf{r=\sqrt{64+225}}}$

$\implies{\sf{r=\sqrt{289}}}$

$\implies{\sf{r=17}}$

$\therefore{\boxed{\sf{r=17}}}$

We know that, (8, 15) lies in 1st quadrant,

$\implies{\sf{Amplitude=tan^{-1}\Bigg(\dfrac{b}{a}\Bigg)}}$

$\implies{\sf{Amplitude=tan^{-1}\Bigg(\dfrac{15}{8}\Bigg)}}$

$\therefore{\boxed{\sf{Amplitude=tan^{-1}\Bigg(\dfrac{15}{8}\Bigg)}}}$

Now, we will convert into Polar form:

$\implies{\sf{z=r(\cos \theta + i\sin \theta)}}$

$\implies{\sf{z=17(\cos \theta + i\sin \theta)}}$

$\therefore{\boxed{\sf{z=17(\cos \theta + i\sin \theta)}}}$

Now, we will convert into exponential form:

$\implies{\sf{z=r\exp^{i\theta}}}$

$\implies{\sf{z=17\exp^{i\theta}}}$

$\therefore{\boxed{\sf{z=17\exp^{i\theta}}}}$