Answer:
-i
Step-by-step explanation:
Given are two complex numbers as
[tex]z1 = \sqrt{2} (cos 135+isin 135)
z2 = \sqrt{2} (cos 315+isin315)[/tex]
To find quotient
We can use Demoivre theorem for products and quotients here
Quotient would be equal to
[tex]\frac{z1}{z2} =\frac{\sqrt{2}(cos45+isin45) }{\sqrt{2}(cos315+isin315) } \\=cos (45-135)+isin(45-135)\\=cos90-isin90\\=-i[/tex]
Hence quotient = -i
Answer with explanation:
If, A complex number, Z=a + i b
then, Z can be written as,=|r| (Cos A + i Sin A), where |r|=Modulus of Complex number which is equal to
[tex]|r|=\sqrt{a^2+b^2}\\\\A=\ Tan^{-1}(\frac{b}{a})\\\\a=r \ Cos A\\\\b=r \ Sin A\\\\Z=|r|e^{iA}[/tex]
The expression which is equivalent to √2[Cos 45° +i Sin 45°] is,
[tex]\Rightarrow \sqrt{2}( \ Cos45^{\circ} + i\ Sin 45^{\circ})\\\\\Rightarrow \sqrt{2}(\frac{1}{\sqrt{2}} +i\frac{1}{\sqrt{2}} )\\\\\Rightarrow \sqrt{2}\times\frac{1+i}{\sqrt{2}}\\\\=1+ i[/tex]
And , the expression which is equivalent to,√2[Cos 315° +i Sin 315°] is
. [tex]\Rightarrow \sqrt{2}( \ Cos315^{\circ} + i\ Sin 315^{\circ})\\\\\Rightarrow \sqrt{2}( \ Cos45^{\circ} - i\ Sin 45^{\circ})\\\\\Rightarrow \sqrt{2}(\frac{1}{\sqrt{2}} -i\frac{1}{\sqrt{2}} )\\\\\Rightarrow \sqrt{2}\times\frac{1-i}{\sqrt{2}}\\\\=1- i[/tex]
→ Cos (360°-45°)=Cos 45°
→Sin (360° -45°)= -Sin 45°
