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Find Re⁡(3i5+i)\operatorname{Re}\left(\frac{3 i}{\sqrt{5}+i}\right).
Jul 1, 2024
Solution by Steps
step 1
To simplify 3i5+i\frac{3i}{\sqrt{5} + i}, multiply the numerator and the denominator by the conjugate of the denominator, 5−i\sqrt{5} - i
step 2
3i5+i⋅5−i5−i=3i(5−i)(5+i)(5−i)\frac{3i}{\sqrt{5} + i} \cdot \frac{\sqrt{5} - i}{\sqrt{5} - i} = \frac{3i(\sqrt{5} - i)}{(\sqrt{5} + i)(\sqrt{5} - i)}
step 3
Simplify the denominator: (5+i)(5−i)=(5)2−(i)2=5−(−1)=6(\sqrt{5} + i)(\sqrt{5} - i) = (\sqrt{5})^2 - (i)^2 = 5 - (-1) = 6
step 4
Simplify the numerator: 3i(5−i)=3i5−3i2=3i5−3(−1)=3i5+33i(\sqrt{5} - i) = 3i\sqrt{5} - 3i^2 = 3i\sqrt{5} - 3(-1) = 3i\sqrt{5} + 3
step 5
Combine the results: 3i5+36=36+3i56=12+i52\frac{3i\sqrt{5} + 3}{6} = \frac{3}{6} + \frac{3i\sqrt{5}}{6} = \frac{1}{2} + \frac{i\sqrt{5}}{2}
step 6
The real part of 12+i52\frac{1}{2} + \frac{i\sqrt{5}}{2} is 12\frac{1}{2}
Answer
The real part is 12\frac{1}{2}.
Key Concept
Simplifying complex fractions
Explanation
To find the real part of a complex fraction, multiply by the conjugate of the denominator and simplify.
Solution by Steps
step 1
To find the polar form of z1=−3+iz_{1} = -3 + i, we first calculate the magnitude rr and the angle θ\theta
step 2
The magnitude rr is given by r=(−3)2+12=9+1=10r = \sqrt{(-3)^2 + 1^2} = \sqrt{9 + 1} = \sqrt{10}
step 3
The angle θ\theta is given by θ=π−tan⁡−1(13)\theta = \pi - \tan^{-1}\left(\frac{1}{3}\right)
step 4
Therefore, the polar form of z1z_{1} is z1=10(cos⁡(π−tan⁡−1(13))+isin⁡(π−tan⁡−1(13)))z_{1} = \sqrt{10} \left(\cos\left(\pi - \tan^{-1}\left(\frac{1}{3}\right)\right) + i \sin\left(\pi - \tan^{-1}\left(\frac{1}{3}\right)\right)\right)
step 5
To find the polar form of z2=3+4iz_{2} = \sqrt{3} + 4i, we first calculate the magnitude rr and the angle θ\theta
step 6
The magnitude rr is given by r=(3)2+42=3+16=19r = \sqrt{(\sqrt{3})^2 + 4^2} = \sqrt{3 + 16} = \sqrt{19}
step 7
The angle θ\theta is given by θ=tan⁡−1(43)\theta = \tan^{-1}\left(\frac{4}{\sqrt{3}}\right)
step 8
Therefore, the polar form of z2z_{2} is z2=19(cos⁡(tan⁡−1(43))+isin⁡(tan⁡−1(43)))z_{2} = \sqrt{19} \left(\cos\left(\tan^{-1}\left(\frac{4}{\sqrt{3}}\right)\right) + i \sin\left(\tan^{-1}\left(\frac{4}{\sqrt{3}}\right)\right)\right)
Answer
The polar form of z1z_{1} is 10(cos⁡(π−tan⁡−1(13))+isin⁡(π−tan⁡−1(13)))\sqrt{10} \left(\cos\left(\pi - \tan^{-1}\left(\frac{1}{3}\right)\right) + i \sin\left(\pi - \tan^{-1}\left(\frac{1}{3}\right)\right)\right).
The polar form of z2z_{2} is 19(cos⁡(tan⁡−1(43))+isin⁡(tan⁡−1(43)))\sqrt{19} \left(\cos\left(\tan^{-1}\left(\frac{4}{\sqrt{3}}\right)\right) + i \sin\left(\tan^{-1}\left(\frac{4}{\sqrt{3}}\right)\right)\right).
Key Concept
Polar Form of Complex Numbers
Explanation
The polar form of a complex number z=x+yiz = x + yi is given by r(cos⁡(θ)+isin⁡(θ))r(\cos(\theta) + i\sin(\theta)), where r=x2+y2r = \sqrt{x^2 + y^2} is the magnitude and θ=tan⁡−1(yx)\theta = \tan^{-1}\left(\frac{y}{x}\right) is the angle.
why
Solution by Steps
step 1
The given equation is θ=π−tan⁡−1(13)θ = π - \tan^{-1}\left(\frac{1}{3}\right)
step 2
To simplify, we start by adding tan⁡−1(13)\tan^{-1}\left(\frac{1}{3}\right) to both sides: θ+tan⁡−1(13)=πθ + \tan^{-1}\left(\frac{1}{3}\right) = π
step 3
Rearrange the equation to isolate θθ: θ=π−tan⁡−1(13)θ = π - \tan^{-1}\left(\frac{1}{3}\right)
step 4
Using the identity for the inverse tangent, we can express θθ as: θ=π−12ilog⁡(1−i3)+12ilog⁡(1+i3)θ = π - \frac{1}{2}i \log\left(1 - \frac{i}{3}\right) + \frac{1}{2}i \log\left(1 + \frac{i}{3}\right)
Answer
θ=π−12ilog⁡(1−i3)+12ilog⁡(1+i3)θ = π - \frac{1}{2}i \log\left(1 - \frac{i}{3}\right) + \frac{1}{2}i \log\left(1 + \frac{i}{3}\right)
Key Concept
Simplification of inverse tangent expression
Explanation
The given equation θ=π−tan⁡−1(13)θ = π - \tan^{-1}\left(\frac{1}{3}\right) can be simplified using the properties of the inverse tangent function and logarithms.
Generated Graph
Solution by Steps
step 1
Convert z1=−3+iz_1 = -3 + i to polar form
step 2
The magnitude ∣z1∣=(−3)2+12=10|z_1| = \sqrt{(-3)^2 + 1^2} = \sqrt{10}
step 3
The argument θ1=tan⁡−1(1−3)+π=π−tan⁡−1(13)\theta_1 = \tan^{-1}\left(\frac{1}{-3}\right) + \pi = \pi - \tan^{-1}\left(\frac{1}{3}\right)
step 4
Therefore, z1=10ei(π−tan⁡−1(1/3))z_1 = \sqrt{10} e^{i(\pi - \tan^{-1}(1/3))}
step 5
Convert z2=3+4iz_2 = \sqrt{3} + 4i to polar form
step 6
The magnitude ∣z2∣=(3)2+42=19|z_2| = \sqrt{(\sqrt{3})^2 + 4^2} = \sqrt{19}
step 7
The argument θ2=tan⁡−1(43)=tan⁡−1(4/3)\theta_2 = \tan^{-1}\left(\frac{4}{\sqrt{3}}\right) = \tan^{-1}(4/\sqrt{3})
step 8
Therefore, z2=19eitan⁡−1(4/3)z_2 = \sqrt{19} e^{i \tan^{-1}(4/\sqrt{3})}
step 9
Calculate (z2)3(z_2)^3: (19eitan⁡−1(4/3))3=193/2ei3tan⁡−1(4/3)(\sqrt{19} e^{i \tan^{-1}(4/\sqrt{3})})^3 = 19^{3/2} e^{i 3 \tan^{-1}(4/\sqrt{3})}
step 10
Divide by z1z_1: 193/2ei3tan⁡−1(4/3)10ei(π−tan⁡−1(1/3))=193/210ei(3tan⁡−1(4/3)−(π−tan⁡−1(1/3)))\frac{19^{3/2} e^{i 3 \tan^{-1}(4/\sqrt{3})}}{\sqrt{10} e^{i(\pi - \tan^{-1}(1/3))}} = \frac{19^{3/2}}{\sqrt{10}} e^{i (3 \tan^{-1}(4/\sqrt{3}) - (\pi - \tan^{-1}(1/3)))}
step 11
Simplify the magnitude: 193/210=1919/10\frac{19^{3/2}}{\sqrt{10}} = 19 \sqrt{19/10}
step 12
Simplify the argument: 3tan⁡−1(4/3)−(π−tan⁡−1(1/3))3 \tan^{-1}(4/\sqrt{3}) - (\pi - \tan^{-1}(1/3))
step 13
Therefore, (z2)3z1=1919/10ei(3tan⁡−1(4/3)−(π−tan⁡−1(1/3)))\frac{(z_2)^3}{z_1} = 19 \sqrt{19/10} e^{i (3 \tan^{-1}(4/\sqrt{3}) - (\pi - \tan^{-1}(1/3)))}
Answer
(z2)3z1=1919/10ei(3tan⁡−1(4/3)−(π−tan⁡−1(1/3)))\frac{(z_2)^3}{z_1} = 19 \sqrt{19/10} e^{i (3 \tan^{-1}(4/\sqrt{3}) - (\pi - \tan^{-1}(1/3)))}
Key Concept
Polar form of complex numbers
Explanation
To convert a complex number to polar form, we need to find its magnitude and argument. The magnitude is the distance from the origin, and the argument is the angle with the positive real axis.
Question iii: Find the square roots of z1z_1 in terms of a+bia + bi.
step 1
Convert z1=−3+iz_1 = -3 + i to polar form
step 2
The magnitude ∣z1∣=(−3)2+12=10|z_1| = \sqrt{(-3)^2 + 1^2} = \sqrt{10}
step 3
The argument θ1=π−tan⁡−1(13)\theta_1 = \pi - \tan^{-1}\left(\frac{1}{3}\right)
step 4
Therefore, z1=10ei(π−tan⁡−1(1/3))z_1 = \sqrt{10} e^{i(\pi - \tan^{-1}(1/3))}
step 5
Find the square roots: z1=104ei(π−tan⁡−1(1/3))/2\sqrt{z_1} = \sqrt[4]{10} e^{i(\pi - \tan^{-1}(1/3))/2} and 104ei(π−tan⁡−1(1/3)+2π)/2\sqrt[4]{10} e^{i(\pi - \tan^{-1}(1/3) + 2\pi)/2}
step 6
Simplify the arguments: π−tan⁡−1(1/3)2\frac{\pi - \tan^{-1}(1/3)}{2} and π−tan⁡−1(1/3)+2π2\frac{\pi - \tan^{-1}(1/3) + 2\pi}{2}
step 7
Convert back to rectangular form: 101/4(cos⁡(π−tan⁡−1(1/3)2)+isin⁡(π−tan⁡−1(1/3)2))10^{1/4} (\cos(\frac{\pi - \tan^{-1}(1/3)}{2}) + i \sin(\frac{\pi - \tan^{-1}(1/3)}{2})) and 101/4(cos⁡(π−tan⁡−1(1/3)+2π2)+isin⁡(π−tan⁡−1(1/3)+2π2))10^{1/4} (\cos(\frac{\pi - \tan^{-1}(1/3) + 2\pi}{2}) + i \sin(\frac{\pi - \tan^{-1}(1/3) + 2\pi}{2}))
step 8
Therefore, the square roots are approximately 0.28485+1.75532i0.28485 + 1.75532i and −0.28485−1.7553i-0.28485 - 1.7553i
Answer
0.28485+1.75532i0.28485 + 1.75532i and −0.28485−1.7553i-0.28485 - 1.7553i
Key Concept
Square roots of complex numbers
Explanation
To find the square roots of a complex number, convert it to polar form, take the square root of the magnitude, and halve the argument. Then convert back to rectangular form.
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