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# Solve for $z:{{z}^{2}}-\left( 3-2i \right)z=\left( 5i-5 \right)$ (a) $z=\left( 2+i \right)and\left( 1+3i \right)$ (b) $z=\left( 2-i \right)and\left( 1-3i \right)$ (c) $z=\left( 2+i \right)and\left( 1-3i \right)$ (d) $z=\left( 2-i \right)and\left( 1+3i \right)$

Last updated date: 09th Aug 2024
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Hint: To solve this question, we will rearrange the given expression. The equation given to us is of the form of a quadratic equation $a{{x}^{2}}+bx+c=0$ , where z is in place of x. We will compare the given equation with the standard quadratic equation and find the values of a, b and c. Once we get those values, we can solve this quadratic equation with any of the available methods to solve a quadratic equation.

We have been given the expression for which we need to find value of $z:{{z}^{2}}-\left( 3-2i \right)z=\left( 5i-5 \right)$ , To find the zeros of this expression, we need to equate it to zero.
${{z}^{2}}-\left( 3-2i \right)z-\left( 5i-5 \right)=0$
Now the above expression can be compared to that of a general quadratic equation, $a{{x}^{2}}+bx+c=0$. Here, the variable is z in place of x. Thus, we will get $a=1,b=-\left( 3-2i \right),c=-\left( 5i-5 \right)$
Therefore, the quadratic equation with us is ${{z}^{2}}-\left( 3-2i \right)z-\left( 5i-5 \right)=0$.
We will solve this equation with the quadratic formula.
Let us apply three values in the quadratic formula,
$z=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
$z=\dfrac{\left( 3-2i \right)\pm \sqrt{{{\left( -\left( 3-2i \right) \right)}^{2}}-4\times 1\times \left( 5i-5 \right)}}{2}$
$=\dfrac{\left( 3-2i \right)\pm \sqrt{\left( {{3}^{2}}-12i+4{{i}^{2}} \right)+4\left( 5i-5 \right)}}{2}$
We know the basic formula,
$\because {{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$ $\because {{i}^{2}}=-1$
$=\dfrac{\left( 3-2i \right)\pm \sqrt{9-12i-4+20i-20}}{2}=\dfrac{\left( 3-2i \right)\pm \sqrt{5+8i-20}}{2}$
$=\dfrac{\left( 3-2i \right)\pm \sqrt{-15+8i}}{2}=\dfrac{\left( 3-2i \right)\pm \sqrt{{{\left( 1+4i \right)}^{2}}}}{2}$
From the above expression, let us take the discriminant $\left( -15+8i \right)$ .
We can write $-15+8i=\left( 1-16 \right)+8i$
1 and 16 are square numbers, thus we can write it as below.
$={{1}^{2}}-{{4}^{2}}+8i={{1}^{2}}+{{\left( 4i \right)}^{2}}+8i$
We got the above expression in the form of the basic identity, ${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$.
$={{\left( 1+4i \right)}^{2}}$
Now substitute this back in the quadratic formula and simplify it.
$\therefore z=\dfrac{\left( 3-2i \right)\pm \sqrt{{{\left( 1+4i \right)}^{2}}}}{2}=\dfrac{\left( 3-2i \right)\pm \left( 1+4i \right)}{2}$
Thus, we got two roots which are,
$\therefore z=\dfrac{\left( 3-2i \right)+\left( 1+4i \right)}{2}$ or $z=\dfrac{\left( 3-2i \right)-\left( 1+4i \right)}{2}$
Now simplify this root further we get the required answer.
$=\dfrac{3-2i+1+4i}{2}$ or $=\dfrac{3-2i-1-4i}{2}$
$=\dfrac{4-2i}{2}=2+i$ or $=\dfrac{2-6i}{2}=1-3i$
$\therefore z=2+i$ or $\therefore z=1-3i$
Hence, we got the value of z as $\left( 2+i \right)$ and $\left( 1-3i \right)$.
$\therefore z=\left( 2+i \right)and\left( 1-3i \right)$ are the required solution.

So, the correct answer is “Option C”.

Note: It is easier to use the quadratic formula, than other methods of solving complex quadratic equations. If we go for the middle term split method, it would be very difficult to identify the right factors and proceeding with wrong factors can lead to confusions and waste of time. It's important that you understand what and how to solve the complex expression by seeing the question itself.