Question

If \[z = 3 - 5i\], then \[{z^3} - 10{z^2} + 54z - 136 = \]________________

Answer 1

Hint: To solve complex numbers, we have to know the value of $i$, and its higher powers, which we can be found from the expression, \[{i^2} = - 1\]

Complete step-by-step answer:
It is given that \[z = 3 - 5i\] , it is given that we have to find the value of \[{z^3} - 10{z^2} + 54z - 136\]
So first let us determine the value of \[{z^3}\]
\[{z^3} = {\left( {3 - 5i} \right)^3}\]
We know the following algebraic identity \[{\left( {a - b} \right)^3} = {a^3} - 3{a^2}b + 3a{b^2} - {b^3}\]
Using the identity in the above equation we get,
\[{z^3} = {3^3} - {3.3^2}\left( {5i} \right) + 3.3{\left( {5i} \right)^2} - {\left( {5i} \right)^3}\]
Let us solve the higher powers in the above equation and substituting the value of $i$,
\[{z^3} = 27 + 125i - 135i - 225\]
\[{z^3} = - 198 - 10i\]
Then let us again find the value of \[{z^2}\]
We will the following algebraic identity to find \[{\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}\]
\[{z^2} = {\left( {3 - 5i} \right)^2}\]
By applying the algebraic identity in the above equation we get,
\[{z^2} = 9 - 2 \times 3 \times 5i + {\left( {5i} \right)^2}\]
\[{z^2} = 9 - 30i - 25 = - 16 - 30i\]
Let us substitute the value of \[{z^3}\], \[{z^2}\] and \[z\] in the given equation whose value has to be found,
That is let us substitute the values in \[{z^3} - 10{z^2} + 54z - 136\], therefore we get
\[{z^3} - 10{z^2} + 54z - 136 = - 198 - 10i - 10\left( { - 16 - 30i} \right) + 54\left( {3 - 5i} \right) - 136\]
Let us now solve the above equation in the right hand side we get
\[{z^3} - 10{z^2} + 54z - 136\]\[ = - 198 - 10i + 160 + 300i + 162 - 270i - 136\]
And again let us solve it so we would find the value of the given equation
\[{z^3} - 10{z^2} + 54z - 136\]\[ = - 12 + 20i\]
Hence we have found the value of the given equation, \[{z^3} - 10{z^2} + 54z - 136 = 20i - 12\].

Note: The general form of the complex number can be expressed as \[x + iy\]
Where \[x\] and \[y\] are real numbers and \[i\]is an imaginary number. We have used the value of higher powers of \[i\], which are noted as follows,
$
  i = \sqrt { - 1} \\
  {i^2} = - 1 \\
  {i^3} = - i \\
  {i^4} = i \\
 $
The value of \[i\] repeats at every fourth power of \[i\].
This number \[i\] came into existence as one cannot find the square root of negative numbers, as \[i\] is just the square root of \[ - 1\].