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Sladkaya [172]
4 years ago
15

Two complex numbers are given byz1= 3 +j4 andz2= 2 +j3. Determine in both polar andCartesian forms

Mathematics
1 answer:
Marat540 [252]4 years ago
3 0

Answer:

Cartesian

z₁= 3 +4*j

z₂= 2 +3*j

Polar

z₁=5 * e^ (0.927*j)

z₁=√13 * e^ (0.982*j)

Step-by-step explanation:

for the complex numbers z the cartesian form of is

z= x + y*j

then

1) z₁= 3 +4*j (cartesian form)

2) z₂= 2 +3*j (cartesian form)

the polar form is

z= r* e^jθ

where

r= √(x²+y²) → r₁ = √(3²+4²) = 5 ,  r₂ = √(2²+3²) = √13

and

θ = tan⁻¹ (y/x)  → θ₁  = tan⁻¹ (4/3)= 0.927 rad , θ₂  = tan⁻¹ (3/2)= 0.982 rad

then

z₁=5 * e^ (0.927*j)

z₁=√13 * e^ (0.982*j)

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Suppose that 35 people are divided in a random manner into two teams in such a way that one team contains 10 people and the othe
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Step-by-step explanation:

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3 years ago
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3 years ago
If tan theta= 3/4 , find csc theta
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Answer:

csc\theta=\frac{5}{3}

Step-by-step explanation:

Given:

tan\theta =\frac{3}{4}

cot\theta =\frac{4}{3}                 [∵ cot\theta =\frac{1}{tan\theta} ]

Squaring both sides.

cot^2\theta =\frac{4^2}{3^2}=\frac{16}{9}

csc^2\theta -1=\frac{16}{9}        [∵ cot^2\theta =csc^2\theta-1] ]

Adding 1 to both sides.

csc^2\theta -1+1=\frac{16}{9}+1

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csc^2\theta=\frac{16+9}{9}

csc^2\theta=\frac{25}{9}

Taking square root both sides.

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