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Math  /  Algebra

Questionb)
Find the two solutions of z2=2−5jz^{2}=2-5 j in the form z1=r1eθ1jz_{1}=r_{1} e^{\theta_{1} j}, (the principal solution), and z2=r2eθ2jz_{2}=r_{2} e^{\theta_{2} j}.
For z1=r1eθ1z_{1}=r_{1} e^{\theta_{1}}, where r1=r_{1}= \qquad and θ1=\theta_{1}= \qquad radians
For z2=r2eθ2jz_{2}=r_{2} e^{\theta_{2} j}, where r2=r_{2}= \qquad and θ2=\theta_{2}= \qquad radians

Studdy Solution

STEP 1

1. We are dealing with complex numbers in polar form.
2. The equation z2=2−5j z^2 = 2 - 5j needs to be solved for z z .
3. The solutions will be expressed in polar form as z1=r1eθ1j z_1 = r_1 e^{\theta_1 j} and z2=r2eθ2j z_2 = r_2 e^{\theta_2 j} .

STEP 2

1. Convert the complex number 2−5j 2 - 5j to polar form.
2. Find the principal square root of the complex number.
3. Find the second square root of the complex number.

STEP 3

First, express the complex number 2−5j 2 - 5j in polar form. This involves finding the magnitude r r and the argument θ \theta .
The magnitude r r is given by:
r=22+(−5)2=4+25=29 r = \sqrt{2^2 + (-5)^2} = \sqrt{4 + 25} = \sqrt{29}
The argument θ \theta is given by:
θ=tan⁡−1(−52) \theta = \tan^{-1}\left(\frac{-5}{2}\right)
Since the complex number is in the fourth quadrant, adjust the angle accordingly:
θ=tan⁡−1(−52)+2π \theta = \tan^{-1}\left(\frac{-5}{2}\right) + 2\pi

STEP 4

To find the principal square root, we need to find z1=r1eθ1j z_1 = r_1 e^{\theta_1 j} .
The magnitude of the square root is:
r1=29=(29)1/2 r_1 = \sqrt{\sqrt{29}} = (\sqrt{29})^{1/2}
The argument of the principal square root is:
θ1=θ2=tan⁡−1(−52)+2π2 \theta_1 = \frac{\theta}{2} = \frac{\tan^{-1}\left(\frac{-5}{2}\right) + 2\pi}{2}

STEP 5

To find the second square root, we need to find z2=r2eθ2j z_2 = r_2 e^{\theta_2 j} .
The magnitude of the second square root is the same as the principal:
r2=r1=(29)1/2 r_2 = r_1 = (\sqrt{29})^{1/2}
The argument of the second square root is:
θ2=θ2+π=tan⁡−1(−52)+2π2+π \theta_2 = \frac{\theta}{2} + \pi = \frac{\tan^{-1}\left(\frac{-5}{2}\right) + 2\pi}{2} + \pi
The solutions are:
For z1=r1eθ1j z_1 = r_1 e^{\theta_1 j} , where r1=(29)1/2 r_1 = (\sqrt{29})^{1/2} and θ1=tan⁡−1(−52)+2π2 \theta_1 = \frac{\tan^{-1}\left(\frac{-5}{2}\right) + 2\pi}{2} radians.
For z2=r2eθ2j z_2 = r_2 e^{\theta_2 j} , where r2=(29)1/2 r_2 = (\sqrt{29})^{1/2} and θ2=tan⁡−1(−52)+2π2+π \theta_2 = \frac{\tan^{-1}\left(\frac{-5}{2}\right) + 2\pi}{2} + \pi radians.

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