Question-115325

Question Number 115325 by zakirullah last updated on 25/Sep/20

Commented by mohammad17 last updated on 25/Sep/20

Q 5 / i)

z^{2} + z + 3 = 0 \Rightarrow z = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} \Rightarrow z = \frac{- 1 \pm i \sqrt{11}}{2}

∴ z = - \frac{1}{2} - \frac{\sqrt{11}}{2} i, z = - \frac{1}{2} + \frac{\sqrt{11}}{2} i

i i) z^{2} - 1 = z \Rightarrow z^{2} - z - 1 = 0 \Rightarrow z = \frac{1 \pm \sqrt{5}}{2} \Rightarrow z = \frac{1}{2} - \frac{\sqrt{5}}{2}, z = \frac{1}{2} + \frac{\sqrt{5}}{2}

i i i) z^{2} - 2 z + i = 0 \Rightarrow z = \frac{2 \pm 2 \sqrt{1 - i}}{2} \Rightarrow z = 1 - \sqrt{1 - i}, z = 1 + \sqrt{1 - i}

i v) z^{2} + 4 = 0 \Rightarrow (z - 2 i) (z + 2 i) = 0 \Rightarrow z = 2 i, z = - 2 i

Commented by zakirullah last updated on 25/Sep/20

solve Q 5, 6

Commented by bobhans last updated on 25/Sep/20

y o u r i m a g e n o t g o o d

Commented by zakirullah last updated on 25/Sep/20

sir click the image on the screen then you can see ?

Commented by mohammad17 last updated on 25/Sep/20

Q 6)

i) z^{4} + z^{2} + 1 = 0

p u t : z^{2} = w \Rightarrow w^{2} + w + 1 = 0 \Rightarrow w = \frac{- 1 \pm i \sqrt{3}}{2} \Rightarrow w = - \frac{1}{2} - \frac{i \sqrt{3}}{2} \Rightarrow z = - \frac{1}{2} + i \frac{\sqrt{3}}{2}, z = \frac{1}{2} - \frac{\sqrt{3}}{2} i

w = - \frac{1}{2} + \frac{\sqrt{3}}{2} i \Rightarrow z = \frac{1}{2} - \frac{\sqrt{3}}{2} i, z = - \frac{1}{2} - \frac{\sqrt{3}}{2} i

i i) {(z - 1)}^{3} = - 1

p u t : m = z - 1 \Rightarrow m^{3} + 1 = 0 \Rightarrow (m + 1) (m^{2} - m + 1) = 0

\Rightarrow m + 1 = 0 \Rightarrow z - 1 + 1 = 0 \Rightarrow z = 0

(m^{2} - m + 1) = 0 \Rightarrow m = \frac{1 \pm \sqrt{3}}{2} i

m = \frac{1}{2} - \frac{\sqrt{3}}{2} i \Rightarrow z - 1 = \frac{1}{2} - \frac{\sqrt{3}}{2} i \Rightarrow z = \frac{3}{2} - \frac{\sqrt{3}}{2} i

m = \frac{1}{2} + \frac{\sqrt{3}}{2} i \Rightarrow z - 1 = \frac{1}{2} + \frac{\sqrt{3}}{2} i \Rightarrow z = \frac{3}{2} + \frac{\sqrt{3}}{2} i