Complex Numbers...

If $${ a }^{ 2 }+{ b }^{ 2 }=1$$, then $$\dfrac {\left( 1+b+ia \right) }{\left( 1+b-ia \right)} $$ is

The real part of $$(1-\cos\theta +2i \sin\theta)^{-1}$$ is?

If in applying the quardratic formula to a quadratic equation
$$f(x) = ax^2 + bx + c = 0$$, it happens that $$c = b^2/4a$$, then the graph of $$y = f(x)$$ will certainly:

If $$\mid{z_1}\mid=2$$, $$\mid{z_2}\mid=3$$, $$\mid{z_3}\mid=4$$ and $$\mid{z_1+z_2+z_3}\mid=2$$, then the value of $$\mid{4z_2z_3+9z_3z_1+16z_1z_2}\mid$$.

$${ \left( \dfrac { cos\dfrac { \pi  }{ 8 } -isin\dfrac { \pi  }{ 8 }  }{ cos\dfrac { \pi  }{ 8 } +isin\dfrac { \pi  }{ 8 }  }  \right)  }^{ 4 }$$

Let $$z_1$$ and $$z_2$$ are two complex numbers such that $$(1-i)z_1=2z_2$$ and $$arg(z_1z_2)=\dfrac{\pi}{2}$$ then $$arg(z_2)$$ is equals to:

Let $$z$$ be a complex number such that $$\left| z+\dfrac { 1 }{ z }  \right| =2$$. 

If $$\left| z \right| ={ r }_{ 1 }$$ and $$\left| \dfrac { 1 }{ z }  \right| =$$ $${r}_{2}$$ for $$\arg z=\dfrac { \pi  }{ 4 }$$ then 

$$\left| { r }_{ 1 }-{ r }_{ 2 } \right| =$$

If $${z_1}$$ and $${z_2}$$ are two non-zero complex number such that $$\left| {{{{z_1}} \over {{z_2}}}} \right|$$ = 2 and $$\arg \left( {{z_1}{z_2}} \right) = {{3\pi } \over 2}$$ , then $${{\overline {{z_1}} } \over {{z_2}}}$$ is equal to 

When will the quadratic equation $$ax^2+bx+c=0$$ have Real Roots?

If the value of '$$b^2-4ac$$'is equal to zero, the quadratic equation $$ax^2+bx+c=0$$ will have