Question

find 2 complex numbers satisfying the equation z^2-2z+4=0​

Answer 1

$\underline{\textbf{Given:}}$

$\mathsf{Equation\;is\;z^2-2z+4=0}$

$\underline{\textbf{To find:}}$

$\textsf{Two complex numbers satisfying the equaion}$

$\mathsf{z^2-2z+4=0}$

$\underline{\textbf{Solution:}}$

$\mathsf{Consider,}$

$\mathsf{z^2-2z+4=0}$

$\textsf{This can be written as}$

$\mathsf{(z^2-2z+1)+3=0}$

$\mathsf{(z^2-2z+1)=-3}$

$\mathsf{(z-1)^2=-3}$

$\mathsf{(z-1)^2=3i^2}$

$\textsf{Taking square root on bothsides, we get}$

$\mathsf{z-1=\pm\,i\sqrt3}$

$\implies\mathsf{z=1\pm\,i\sqrt3}$

$\therefore\textbf{Two complex numbers satisfying the equation is}$

$\;\bf\,1+i\sqrt3\;\;and\;\;1-i\sqrt3$

Answer 2

Answer:

Two complex roots of $z^{2} - 2z + 4 =0$ are $z = 1$ ± $i\sqrt{3}$

Step-by-step explanation:

Given:- $z^{2} - 2z + 4 =0$

To Find:- Two complex numbers satisfying the above equation.

Solution:-  $z^{2} - 2z + 4 =0$

The above equation can also be written as:

⇒ $z^{2} - 2z + 1 + 3 =0$

⇒ $z^{2} - 2z + 1 = -3$

⇒ $(z-1)^{2} = 3i^{2}$

Now, taking square root on both sides

⇒ $\sqrt{(z-1)^{2} } = \sqrt{3i^{2} }$

⇒ $z-1 = i\sqrt{3}$

⇒ $z = 1$ ± $i\sqrt{3}$

∴ Two complex roots of $z^{2} - 2z + 4 =0$ are $z = 1$ ± $i\sqrt{3}$

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