The polar form of complex number z - i-1-cos-3-isin-3 is-

The correct option is A √(2)(cos5π/12+i sin5π/12) z= #10; i 1 cosπ #160; 3 #160; +isinπ #10; #160; 3 #160; #160; z= i 1 e ^ iπ ...

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Standard XII

Mathematics

Domain and Range of Basic Inverse Trigonometric Functions

The polar for...

Question

The polar form of complex number z $=$ $\frac{i - 1}{cos \frac{π}{3} + i sin \frac{π}{3}}$ is.

\frac{1}{\sqrt{2}} (cos \frac{3 π}{12} + i sin \frac{3 π}{12})

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\sqrt{2} (cos \frac{5 π}{12} + i sin \frac{5 π}{12})

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\sqrt{2} (cos \frac{7 π}{12} + i sin \frac{7 π}{12})

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\frac{1}{\sqrt{2}} (cos \frac{5 π}{12} + i sin \frac{5 π}{12})

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Solution

The correct option is A $\sqrt{2} (cos \frac{5 π}{12} + i sin \frac{5 π}{12})$
$z = \frac{i - 1}{cos \frac{π}{3} + i sin \frac{π}{3}} z = \frac{i - 1}{e^{\frac{i π}{3}}} i - 1 = \frac{\sqrt{2} (- 1 + i)}{\sqrt{2}} = \sqrt{2} (\frac{- 1}{\sqrt{2}} + i \frac{1}{\sqrt{2}})$
$i - 1 = \sqrt{2} (cos \frac{3 π}{4} + i sin \frac{3 π}{4}) = \sqrt{2} e^{\frac{i 3 π}{4}}$
$z = \sqrt{2} \frac{e^{\frac{i 3 π}{4}}}{e^{\frac{i π}{3}}} = \sqrt{2} e^{i ⎛ ⎝ \frac{3 π}{4} - \frac{π}{3} ⎞ ⎠}$
$z = \sqrt{2} e^{\frac{i 5 π}{12}} z = \sqrt{2} (cos \frac{5 π}{12} + i sin \frac{5 π}{12})$
So option $B$ is correct

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Similar questions

Q. Let

z_{1} = 1 - i

and

z_{2} = \sqrt{3} + i

, then

z_{1} \cdot z_{2}

in polar form is

Find all the cube root of $- 4 \sqrt{2} - 4 \sqrt{2} i$

{⎡ ⎢ ⎢ ⎢ ⎢ ⎣ \frac{1 + cos (\frac{π}{12}) + i sin (\frac{π}{12})}{1 + cos (\frac{π}{12}) - i sin (\frac{π}{12})} ⎤ ⎥ ⎥ ⎥ ⎥ ⎦}^{72} =

Prove that:
(i) $2 s i n \frac{5 π}{12} s i n \frac{π}{12} = \frac{1}{2}$
(ii) $2 c o s \frac{5 π}{12} c o s \frac{π}{12}$
(iii) $2 s i n \frac{5 π}{12} c o s \frac{5 π}{12} = \frac{\sqrt{3} + 2}{2}$

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The polar form of complex number z = i−1cosπ3+isinπ3 is.

The correct option is A √2(cos5π12+isin5π12)z=i−1cosπ3+isinπ3z=i−1eiπ3i−1=√2(−1+i)√2=√2(−1√2+i1√2)i−1=√2(cos3π4+isin3π4)=√2ei3π4z=√2ei3π4eiπ3=√2ei⎛⎝3π4−π3⎞⎠ z=√2ei5π12z=√2(cos5π12+isin5π12)So option B is correct

The polar form of complex number z $=$ $\frac{i - 1}{cos \frac{π}{3} + i sin \frac{π}{3}}$ is.