Question

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

Answer 1

Answer:

6i³+7i⁴+5i³+3i²-4=-11i

square root of the answer = -121

Step-by-step explanation:

6i³+7i⁴+5i³+3i²-4

=-6i+7-5i-3-4 (i³=-i ,i²=-1, i⁴=1)

=0-11i

=-11i

square root of -11i=(-11i)²=121i²=-121

Answer 2

Step-by-step explanation:

i square is equal to -1.

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

3

+7i

4

+5i

3

+3i

2

−4=>

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

2

+5(−1)i+3(−1)

- 4 = > - 11i + 7 - 3 - 4 = > - 11i−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} } = >−11i=11(−i)=11e

i

2

3π

=>

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}}

11

e

i

2×2

3π

and−

11

e

i

2×2

3π

So the square roots of (-11i) are

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

11

e

i

4

3π

and−

11

e

i

4

3π

Hope it helps you.