Answer:
[tex] \large \boxed{ \boxed{ \red{ \ \rm 5 \sqrt{2} ( \cos( \dfrac{7\pi}{4} ) + i \sin( \dfrac{7\pi}{4} ) )}}}[/tex]
Step-by-step explanation:
to understand this
you need to know about:
tips and formulas:
- [tex] \sf \: formula \: of \: trigonometric \: form : \\ \rm r( \cos( \theta) + i \sin( \theta) )[/tex]
- [tex] \rm \: r = \sqrt{ {a}^{2} + {b}^{2} } [/tex]
- [tex] \rm \theta = \arctan( \dfrac{b}{a} )[/tex]
given:
let's solve:
complex number:a+bi
therefore
[tex] \sf sustitute \: the \: value \: of \: a \: and \: b \: into \: the \: modulus : [/tex]
[tex] \rm \: r = \sqrt{ {5}^{2} + ( { - 5)}^{2} } [/tex]
[tex] \sf simplify \: squres : [/tex]
[tex] \rm \: r = \sqrt{25 + 25} [/tex]
[tex] \sf simplify \: addition : [/tex]
[tex] \rm \: r = \sqrt{50} [/tex]
[tex] \sf rewrite \: 50 \: as \: 2 \times 25[/tex]
[tex] \sf r = \sqrt{2 \times 25} [/tex]
[tex] \sf use \: \sqrt{ab} \iff \: \sqrt{a } \sqrt{b} : [/tex]
[tex] \rm r = \sqrt{25} \sqrt{2} [/tex]
[tex] \therefore \rm \: r = 5 \sqrt{2} [/tex]
[tex] \sf sustitute \: the \: value \: of \: a \: and \: b \: into \: the \: agrument : [/tex]
[tex] \rm \theta = \arctan(\dfrac{ - 5}{5} )[/tex]
[tex] \sf simplify \: divition : [/tex]
[tex] \rm \theta = \arctan( - 1)[/tex]
[tex] \theta = - \dfrac{\pi}{4} [/tex]
[tex] \sf add \: 2\pi \: to \: get \: a \: positive \: agrument[/tex]
[tex] \rm \theta = - \dfrac{\pi}{4} + 2\pi \\ \rm\therefore \: \theta = \frac{7\pi}{4} [/tex]
[tex] \sf \: sustitute \: the \: value \: of \: \rm \: r \: and \: \theta : \\ \rm 5 \sqrt{2} ( \cos( \frac{7\pi}{4} ) + i \sin( \frac{7\pi}{4} ) )[/tex]
[tex]\text{And we are done!}[/tex]
