Question

find the modulus of (1-i)^10​

Answer 1

Answer:

hope you understand!!!!

Answer 2

Question :-

$\sf \: Find \: the \: modulus \: of \: {(1 - \iota)}^{10}$

Answer :-

$\mapsto \red{ \boxed{ \bf \: - 32 \: \iota \green{ \: or \: - 32 \sqrt{ - 1} }}} \:$

Solution :-

We have ,

$\longmapsto \bf \: {(1 - \iota)}^{10}$

Split the power

$\mapsto \bf { \{{(1 - \iota)}^{2} \}}^{5} \\ \\ \: \: \: \: \: \: \: \: \because \orange{ \boxed{\sf {(x + y)}^{2} = {x}^{2} + {y}^{2} + 2xy}} \\ \therefore \: \\ \mapsto \bf \: {(1 + { \iota}^{2} - 2 \iota) }^{5} \\ \\ \because \red{\boxed{\sf \: \iota = \sqrt{ - 1} }} \\ \\ \mapsto \bf \: {(1 - 1 - 2 \iota)}^{5} \\ \\ \mapsto \bf {( - 2 \iota)}^{5} \\ \\ \because \boxed{\bf \iota = \sqrt{ - 1} \: so \: { \iota}^{4} = 1 } \\ \\ \mapsto \bf \: {( - 2)}^{5} { \iota}^{4} \: . \iota \\ \\ \mapsto \red{ \boxed{ \bf \: - 32 \: \iota \green{ \: or \: - 32 \sqrt{ - 1} }}}$