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Write the complex number in the form a + bi. 4[cos (-135°) + i sin (-135°)]

User GlinesMome
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so hmmm notice the picture below, is a negative angle, it lands on the 45° of the 3rd quadrant, where "x" and "y" are both negative, and also are both the same value

thus


\bf \begin{array}{llll} 4[cos(&-135^o)+i\ sin(&-135^o)]\\ \uparrow &\quad \uparrow &\quad \uparrow \\ r&\quad \theta&\quad \theta \end{array}\qquad \begin{cases} x=rcos(\theta)\\ y=rsin(\theta)\\ ----------\\ r=4\\ \theta=-135^o\\ cos(-135^o)=-(√(2))/(2)\\ sin(-135^o)=-(√(2))/(2)\\ \end{cases} \\\\\\ x=4\left( -(√(2))/(2) \right)\implies x=-2√(2) \\\\\\ y=4\left( -(√(2))/(2) \right)\implies y=-2√(2)\\\\ -----------------------------\\\\


\bf \begin{array}{llll} a&+&bi\\ x&&yi \end{array}\implies -2√(2)-2√(2)\ i
Write the complex number in the form a + bi. 4[cos (-135°) + i sin (-135°)]-example-1
User Alfonso Tesauro
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