Search

You can find the results of your search below.

Question 9, Exercise 1.2
3 Hits, Last modified:
{(\sqrt{13})^4}\\ &= \frac{12}{169}. \end{align} Therefore, the real part is \(\dfrac{5}{169}\) and the ... c{-7 - 6}{49 + 4} = \frac{-13}{53}. \end{align} Therefore, the real part is \(\dfrac{19}{53}\) and the ... right]}{1681}\\ =& \frac{-720}{1681} \end{align} Therefore, the real part of \(\left(\dfrac{5 - 4i}{5 +
Question 3, Exercise 1.3
3 Hits, Last modified:
} \\ &= \dfrac{1 \pm \sqrt{271}i}{4} \end{align} Therefore, the solution set is: $\left\{\dfrac{1 \pm \s... \dfrac{{6 \pm 8i}}{2} \\ &= 3 \pm 4i \end{align} Therefore, the solution set is: $\{3 \pm 4i\}$ ====Que... } \\ &= \dfrac{{9 \pm \sqrt{37}}}{2} \end{align} Therefore, the solution set $=\left\{ \dfrac{9 \pm \sqr
Question 10, Exercise 1.2
1 Hits, Last modified:
n} Since \[\sqrt{5} \leq \sqrt{13} + \sqrt{10},\] therefore, from (i) and (ii), we have \[\left|z_1 + z_
Question 1, Exercise 1.4
1 Hits, Last modified:
pi - \frac{\pi}{4} = \frac{3\pi}{4}. \end{align} Therefore, the polar form of the complex number \( i -
Question 8, Review Exercise
1 Hits, Last modified:
i \sqrt{2}= \\ \implies & x_{\max}=2. \end{align} Therefore, the amplitude is $2$nm. ====Go to ==== <te