Let z 1-1- i and z 2-3- i- then z 1- z 2 in polar form isA- 2cos-6- i sin-6B- 2 -2cos5 -12- i sin5 -12C- 2cos-12- i sin-12D- 2 -2cos-12- i sin-12

Z_1 = 1 i and z_2 = √(3)+i ∴ z_1 · z_2=( 1 i)( √(3)+i) ⇒ z_1 · z_2=√(3)+i √(3)i+1 ⇒ z_1 · z_2=(√(3)+1) i(√(3) 1) |z_1· z_2|=√((√(3)+1)^2+(√(3) 1)^2)=2√(2) Now ...

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Standard XII

Mathematics

Geometrical Representation of a Complex Number

Let z 1=1- i ...

Question

Let $z_{1} = 1 - i$ and $z_{2} = \sqrt{3} + i$ , then $z_{1} \cdot z_{2}$ in polar form is

2 (cos \frac{π}{6} + i sin \frac{π}{6})

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

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

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

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Solution

The correct option is D $2 \sqrt{2} (cos (- \frac{π}{12}) + i sin (- \frac{π}{12}))$
$z_{1} = 1 - i$ and $z_{2} = \sqrt{3} + i$

$∴ z_{1} \cdot z_{2} = (1 - i) (\sqrt{3} + i) \Rightarrow z_{1} \cdot z_{2} = \sqrt{3} + i - \sqrt{3} i + 1 \Rightarrow z_{1} \cdot z_{2} = (\sqrt{3} + 1) - i (\sqrt{3} - 1)$
$| z_{1} \cdot z_{2} | = \sqrt{(\sqrt{3} + 1)^{2} + (\sqrt{3} - 1)^{2}} = 2 \sqrt{2}$

Now, $tan α = \frac{| \sqrt{3} - 1 |}{| \sqrt{3} + 1 |} \Rightarrow α = \frac{π}{12}$
$∵ z_{1} \cdot z_{2}$ lies in fourth quadrant,
$z_{1} \cdot z_{2} = 2 \sqrt{2} (cos (- α) + i sin (- α))$
$z_{1} \cdot z_{2} = 2 \sqrt{2} (cos (- \frac{π}{12}) + i sin (- \frac{π}{12}))$

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Geometrical Representation of a Complex Number

Standard XII Mathematics

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Let z1=1−i and z2=√3+i, then z1⋅z2 in polar form is

Let $z_{1} = 1 - i$ and $z_{2} = \sqrt{3} + i$ , then $z_{1} \cdot z_{2}$ in polar form is