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

If z=3−5iz = 3 - 5i, then z3−10z2+54z−136={z^3} - 10{z^2} + 54z - 136 = ________________

Explanation

Solution

To solve complex numbers, we have to know the value of ii, and its higher powers, which we can be found from the expression, i2=−1{i^2} = - 1

Complete step-by-step answer:
It is given that z=3−5iz = 3 - 5i , it is given that we have to find the value of z3−10z2+54z−136{z^3} - 10{z^2} + 54z - 136
So first let us determine the value of z3{z^3}
z3=(3−5i)3{z^3} = {\left( {3 - 5i} \right)^3}
We know the following algebraic identity (a−b)3=a3−3a2b+3ab2−b3{\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,
z3=33−3.32(5i)+3.3(5i)2−(5i)3{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 ii,
z3=27+125i−135i−225{z^3} = 27 + 125i - 135i - 225
z3=−198−10i{z^3} = - 198 - 10i
Then let us again find the value of z2{z^2}
We will the following algebraic identity to find (a−b)2=a2−2ab+b2{\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}
z2=(3−5i)2{z^2} = {\left( {3 - 5i} \right)^2}
By applying the algebraic identity in the above equation we get,
z2=9−2×3×5i+(5i)2{z^2} = 9 - 2 \times 3 \times 5i + {\left( {5i} \right)^2}
z2=9−30i−25=−16−30i{z^2} = 9 - 30i - 25 = - 16 - 30i
Let us substitute the value of z3{z^3}, z2{z^2} and zz in the given equation whose value has to be found,
That is let us substitute the values in z3−10z2+54z−136{z^3} - 10{z^2} + 54z - 136, therefore we get
z3−10z2+54z−136=−198−10i−10(−16−30i)+54(3−5i)−136{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, z3−10z2+54z−136=20i−12{z^3} - 10{z^2} + 54z - 136 = 20i - 12.

Note: The general form of the complex number can be expressed as x+iyx + iy
Where xx and yy are real numbers and iiis an imaginary number. We have used the value of higher powers of ii, which are noted as follows,
i=−1 i2=−1 i3=−i i4=i  i = \sqrt { - 1} \\\ {i^2} = - 1 \\\ {i^3} = - i \\\ {i^4} = i \\\
The value of ii repeats at every fourth power of ii.
This number ii came into existence as one cannot find the square root of negative numbers, as ii is just the square root of −1 - 1.