Complex Numbers and Quadratic Equations Test -1

Explanation:

Purely imaginary number is a complex number which has only imaginary part ( iy) 

But if  y=0 the complex number iy will become 0 which is real.

Hence the condition for a number to be purely imaginary is x=0 and y≠0

 

$$x^{2}-px+p+3=0$$ 

Let $$z = r(cos\theta +isin \theta ) $$

$$|7-5i| = \sqrt {54 } $$

$$\begin{array}{l} we\, \, write\, , \\ { \left| z \right| ^{ n } }={ \left| z \right| ^{ n-2 } }\left( { { z^{ 2 } }+z } \right) +1\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \left\{ { we\, know,\, \, all\, \, are\, real\, no. } \right.  \\ so,\, \, \, \, { z^{ 2 } }+z={ \overline { z } ^{ 2 } }+\overline { z } \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \left[ { where\, ,\, \, { z^{ 2 } }+z\, \, \, \, is\, real\, no.\,  } \right.  \\ \Rightarrow { z^{ 2 } }\, -{ \overline { z } ^{ 2 } }+z-\overline { z } =0 \\ \Rightarrow (z-\overline { z } )\, \, (z+\overline { z } +1)=0\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \left| \begin{array}{l} \, z=x+i\, y \\ \, \overline { z } =x-i\, y \end{array} \right.  \\ \, \, \therefore \, \, \, \, 2\, i\, y\, (2x+1)=0 \\ \, \, \, y=0\, \, \, \, \, \, \, \, \, \, \, or\, \, \, \, \, \, x=-\frac { 1 }{ 2 } \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, (Here,\, \, \, z\, \, is\, a\, \, \, real\, \, \, number. \\ Let\, see, \\ \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \left| z \right| =x \\ \left| \begin{array}{l} \, \, { x^{ n } }={ x^{ n-2 } }.{ x^{ 2 } }+1 \\ \, \, \, { \left| x \right| ^{ n-2 } }.\, x=-1 \end{array} \right.  \\ \, \, \, \therefore \, \, \, \, x=-1\,  \\ so,\, number\, of\, z\, =1\, \, and\, \, the\, correct\, option\, is\, B.\, \, \, \, \,  \end{array}$$

$$z_1=3+4i\\z_2=4-5i\\z_1+z_2=3+4i+4-5i\\7-i$$

$$z_1=\sqrt { 3 } -i,z_2=1+i\sqrt { 3 } $$ 

Given $$z_1=3+4i\\z_2=2-i\\z_2-z_1=2-i-3-4i=-1-5i$$

Let $$P={ [\cfrac { 1+i }{ 1-i } ] }^{ 2013 }={ [\cfrac { \sqrt { 2 } { e }^{ \cfrac { i\pi  }{ 4 }  } }{ \sqrt { 2 } { e }^{ \cfrac { -i\pi  }{ 4 }  } } ] }^{ 2013 }=({ e }^{ \cfrac { i2013\pi  }{ 2 }  })$$

$$z_1=8+9i\\z_2=4-6i\\z_1-z_2=8+9i-4+6i=4+15i$$

Let $$z=x+iy\\|z|=\sqrt{x^2+y^2}=1\\x^2+y^2=1\\\dfrac 1z=\dfrac1{x+iy}\\\dfrac{x-iy}{\sqrt{x^2+y^2}}=x-iy\\\left|\dfrac 1z\right|=\sqrt{x^2+y^2}=1$$