Question

6i^3+7i^4+5i^3+3i^2-4 in the form of a+ib and find its square root​

Answer 1

Step-by-step explanation:

i square is equal to -1.

$6 {i}^3 + 7 {i}^4 + 5 {i}^3 + 3 {i}^{2} - 4 = >$

$6( - 1)i + 7 {( - 1)}^{2} + 5( - 1)i + 3( - 1)$

$- 4 = > - 11i + 7 - 3 - 4 = > - 11i$

we can express -11i as 0-11i in the form of a+ib . Let's express -11i in Euler's form.

$- 11i = 11( - i) = 11 {e}^{i \frac{3\pi}{2} } = >$

Apply square root now we get

$\sqrt{11} {e}^{i \frac{3\pi}{2 \times 2} } \: and \: - \sqrt{11} {e}^{i \frac{3\pi}{2 \times 2}}$

So the square roots of (-11i) are

$\sqrt{11} {e}^{i \frac{3\pi}{4} } \: and \: - \sqrt{11} {e}^{ i \frac{3\pi}{4}}$

Hope it helps you.