Question

Find the modulus of ${\left( {1 - i} \right)^{10}}$.

Answer 1

Hint: Modulus of the complex number is the space of the factor on the argand plane representing the complex quantity z from the origin. The modulus of an equation is found out by taking the square root of the sum of the squares of the real part of the terms, i.e., if we have an equation of form $a + ib$, then its modulus will be $\left| {a + ib} \right| = \sqrt {{a^2} + {b^2}} $.

Complete step-by-step solution:
The given equation is: ${\left( {1 - i} \right)^{10}}$
The equation $1 - i$ is of the form $a + ib$ where $a = 1$ and $b = - 1$
The modulus is given by $\left| {{{\left( {1 - i} \right)}^{10}}} \right| = {\left| {1 - i} \right|^{10}}$
Using the modulus formula $\left| {a + ib} \right| = \sqrt {{a^2} + {b^2}} $, we substitute these values,
$\left| {\left( {1 - i} \right)} \right| = \sqrt {{{(1)}^2} + {{( - 1)}^2}} $
Squaring the terms,
$\left| {\left( {1 - i} \right)} \right| = \sqrt {1 + 1} $
Adding the terms,
$\left| {\left( {1 - i} \right)} \right| = \sqrt 2 $
Taking power of $10$ on both sides,
\[{\left| {\left( {1 - i} \right)} \right|^{10}} = {\left( {\sqrt 2 } \right)^{10}}\]
We can write the square root as,
\[{\left| {\left( {1 - i} \right)} \right|^{10}} = {2^{\dfrac{1}{2} \times 10}}\]
Dividing the terms in power,
\[{\left| {\left( {1 - i} \right)} \right|^{10}} = {2^5}\]
Solving the power term,
\[{\left| {\left( {1 - i} \right)} \right|^{10}} = 2 \times 2 \times 2 \times 2 \times 2 = 32\]
Therefore, the modulus of ${\left( {1 - i} \right)^{10}}$ is $32$.

Note: The modulus in the above problem can be found out by an alternative approach also. We can write ${\left( {1 - i} \right)^{10}}$in power of $5$ i.e., ${[{(1 - i)^2}]^5}$ and by expanding the square term using ${(a - b)^2} = {a^2} - 2ab + {b^2}$ we get, ${(1 - 2i - 1)^5}$, on subtracting we get, ${( - 2i)^5}$which gives us $32{( - i)^5} = 32$.