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 Certainly! Here are 50 challenging problems involving complex numbers, along with their solutions. These problems cover a range of topics including algebra, geometry, and functions involving complex numbers.


### Problems and Solutions


1. **Find the modulus and argument of the complex number \( z = 1 + i \).**


   **Solution:**  

   Modulus: \( |z| = \sqrt{1^2 + 1^2} = \sqrt{2} \).  

   Argument: \( \arg(z) = \tan^{-1}\left(\frac{1}{1}\right) = \frac{\pi}{4} \).


2. **Find the roots of the equation \( z^2 + 1 = 0 \).**


   **Solution:**  

   \( z^2 = -1 \).  

   \( z = \pm i \).


3. **Compute \( (2 + 3i)(1 - i) \).**


   **Solution:**  

   \( (2 + 3i)(1 - i) = 2 - 2i + 3i - 3i^2 = 2 + i + 3 = 5 + i \).


4. **If \( z = 2 + 3i \), find \( \frac{1}{z} \).**


   **Solution:**  

   \( \frac{1}{z} = \frac{1}{2 + 3i} \times \frac{2 - 3i}{2 - 3i} = \frac{2 - 3i}{4 + 9} = \frac{2 - 3i}{13} = \frac{2}{13} - \frac{3i}{13} \).


5. **Find the complex conjugate of \( z = 4 - 5i \).**


   **Solution:**  

   The complex conjugate is \( \overline{z} = 4 + 5i \).


6. **Solve for \( z \) in \( z^3 = 8 \).**


   **Solution:**  

   \( 8 = 8 \text{cis} 0 \).  

   The cube roots are \( 2 \text{cis} \left(\frac{2k\pi}{3}\right) \) for \( k = 0, 1, 2 \).  

   So, \( z = 2, -1 + i\sqrt{3}, -1 - i\sqrt{3} \).


7. **Determine the argument of \( z = -1 - i \).**


   **Solution:**  

   \( \arg(z) = \tan^{-1}\left(\frac{-1}{-1}\right) = \frac{3\pi}{4} \) (since it's in the third quadrant).


8. **Find the real and imaginary parts of \( z = \frac{2 - i}{1 + i} \).**


   **Solution:**  

   \( \frac{2 - i}{1 + i} \times \frac{1 - i}{1 - i} = \frac{(2 - i)(1 - i)}{1 + 1} = \frac{2 - 2i - i + i^2}{2} = \frac{1 - 3i}{2} = \frac{1}{2} - \frac{3i}{2} \).


9. **Find \( z \) if \( z^4 = 16 \).**


   **Solution:**  

   \( 16 = 16 \text{cis} 0 \).  

   The fourth roots are \( 2 \text{cis} \left(\frac{k\pi}{2}\right) \) for \( k = 0, 1, 2, 3 \).  

   So, \( z = 2, -2, 2i, -2i \).


10. **Find the magnitude of \( \frac{1+i}{2-i} \).**


    **Solution:**  

    \( \frac{1+i}{2-i} \times \frac{2+i}{2+i} = \frac{(1+i)(2+i)}{4 + 1} = \frac{2 + 3i - 1}{5} = \frac{1 + 3i}{5} \).  

    Magnitude: \( \left|\frac{1 + 3i}{5}\right| = \frac{|1 + 3i|}{5} = \frac{\sqrt{1^2 + 3^2}}{5} = \frac{\sqrt{10}}{5} = \frac{\sqrt{10}}{5} \).


11. **If \( z = 2 + 2i \), find \( z^2 \).**


    **Solution:**  

    \( z^2 = (2 + 2i)^2 = 4 + 8i - 4 = 8i \).


12. **Find \( \text{arg}(2 + 2i) \).**


    **Solution:**  

    \( \text{arg}(2 + 2i) = \tan^{-1}(1) = \frac{\pi}{4} \).


13. **Find the distance between \( z_1 = 1 + i \) and \( z_2 = 3 + 4i \).**


    **Solution:**  

    Distance: \( |z_2 - z_1| = |(3 + 4i) - (1 + i)| = |2 + 3i| = \sqrt{2^2 + 3^2} = \sqrt{13} \).


14. **Solve \( z^2 - 4z + 8 = 0 \).**


    **Solution:**  

    Use the quadratic formula:  

    \( z = \frac{4 \pm \sqrt{16 - 32}}{2} = \frac{4 \pm \sqrt{-16}}{2} = \frac{4 \pm 4i}{2} = 2 \pm 2i \).


15. **Evaluate \( \left(1 + \frac{\sqrt{3}}{2}i\right) \cdot \left(1 - \frac{\sqrt{3}}{2}i\right) \).**


    **Solution:**  

    \( \left(1 + \frac{\sqrt{3}}{2}i\right) \cdot \left(1 - \frac{\sqrt{3}}{2}i\right) = 1 - \left(\frac{\sqrt{3}}{2}\right)^2 i^2 = 1 - \frac{3}{4}(-1) = 1 + \frac{3}{4} = \frac{7}{4} \).


16. **Find the sum and product of the complex numbers \( 1 + i \) and \( 2 - i \).**


    **Solution:**  

    Sum: \( (1 + i) + (2 - i) = 3 \).  

    Product: \( (1 + i)(2 - i) = 2 - i + 2i - i^2 = 2 + i + 1 = 3 + i \).


17. **Find the value of \( (1+i)^6 \).**


    **Solution:**  

    \( 1+i = \sqrt{2} \text{cis} \frac{\pi}{4} \).  

    \( (1+i)^6 = (\sqrt{2} \text{cis} \frac{\pi}{4})^6 = (\sqrt{2})^6 \text{cis} \frac{6\pi}{4} = 8 \text{cis} \frac{3\pi}{2} = -8i \).


18. **Find \( z \) if \( z^2 = -5 + 12i \).**


    **Solution:**  

    Write \( -5 + 12i \) in polar form:  

    \( -5 + 12i = 13 \text{cis} \theta \) where \( \theta = \tan^{-1}\left(\frac{12}{-5}\right) \).  

    The square roots are \( \sqrt{13} \text{cis} \left(\frac{\theta}{2} + k\pi\right) \) for \( k = 0, 1 \).  

    So, \( z = \sqrt{13} \text{cis} \frac{\theta}{2}, -\sqrt{13} \text{cis} \left(\frac{\theta}{2} + \pi\right) \).



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