Question

express into polar form​

Answer 1

Question :

$\frac{1 +i}{( {2 - i})^{2} }$

Answer:

$let \\ \\ z = \frac{1 + 7i}{( {2 - i)}^{2} } \\ \\ \implies \frac{1 + 7i}{ {2}^{2} + {i}^{2} - 2(2)i} \\ \\ \implies \frac{1 + 7i}{4 - 1 - 4i} \\ \\ \implies \: \frac{1 + 7i}{3 - 4i}$

Now Rationalise,

$\implies \: \frac{1 + 7i}{3 + 4i} \times \frac{3 + 4i}{3 + 4i} \\ \\ \implies \: \frac{1(3 + 4i) + 7i(3 + 4i)}{ {3}^{2} - {4}^{2} (- {i}^{2}) } \\ \\ \implies \frac{3 + 4i + 21i + 28( - 1)}{9 + 16} \\ \\ \huge \boxed{\because { - i}^{2} = - 1} \\ \\ \implies \frac{3 - 28 - 25i}{25} \\ \\ \implies \frac{ - 25 + 25i}{25} \\ \\ \implies \frac{ - 25}{25} + \frac{25i}{25} \\ \\ \implies - 1 + i$

Let,

$- 1 + i = r(cos \theta + i \: sin \theta)$

By equating real and imaginary part,

$r \: cos \theta = - 1 \: .........(1) \\ \\ r \: sin \theta = 1 \: \: ...........(2)$

By adding and squaring both side we get,

${r}^{2} ( {cos}^{2} \theta + {sin}^{2} \theta) = ({ - 1})^{2} + {(1)}^{2} \\ \\ {r}^{2} (1) = 2$

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$\because \: {cos}^{2} \theta + {sin}^{2} \theta = 1$

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$\implies {r}^{2} = 2 \\ \\ \implies \: r = \sqrt{2}$

Put the value of root 2 in (1) and (2) equation,

from (1)

$r cos \theta \: = 1 \\ \\ \sqrt{2} cos \theta = 1 \\ \\ cos \theta = \frac{ - 1}{ \sqrt{2} }$

From (2),

$r \: sin \theta = 1 \\ \\ \sqrt{2} sin \theta = 1 \\ \\ sin \theta = \frac{1}{ \sqrt{2} }$

Now

$\huge\boxed{ \theta \: is \: in \: the \: 2nd \: quadrant \: and \: - \pi < \theta \leqslant \pi}$

$\boxed{ \therefore \theta = \frac{3\pi}{4} } \\ \\ \therefore - 1 + i = \sqrt{2} (cos \frac{3\pi}{4} + i \: sin \frac{3\pi}{4}$

Hence

$z \: = \frac{ {1 + 7i} }{ {(1 - 2i)}^{2} } \: is \: represented \:by \: p( \sqrt{2} \: \: \: \frac{3\pi}{4} )$

Answer 2

Answer:

Polar form is that it is not completely donate it's electro for ex HCl

Step-by-step explanation: