Number Theory T...

If $${z}_{1},{z}_{2}$$ are two complex numbers and $$c>0$$ such that $${ \left| { z }_{ 1 }+{ z }_{ 2 } \right|  }^{ 2 }\le \left( 1+c \right) { \left| { z }_{ 1 } \right|  }^{ 2 }+k{ \left| { z }_{ 2 } \right|  }^{ 2 },$$ then $$k=$$

Find the product and write the answer in standard form.
$$\left( 2-4i \right) \left( 3+7i \right) $$

The product of $$(3-2i)$$ and $$\left(\dfrac { 5 }{ 2 } -4i\right)$$, if $$i=\sqrt { -1 } $$ , is:

The resultant complex number when $$(4+6i)$$ is divided by $$(10-5i)$$ is

Find the product. Write the answer in standard form.
$$i\left( 6-2i \right) \left( 7-5i \right) $$

Let $${ X }_{ n }=\left\{ z=x+iy:{ \left| z \right|  }^{ 2 } \le \dfrac { 1 }{ n }  \right\} $$ for all integers $$n\ge 1$$. Then, $$\displaystyle\bigcap _{ n=1 }^{ \infty  }{ { X }_{ n } } $$ is

The modulus of $$\dfrac { 1-i }{ 3+i } +\dfrac { 4i }{ 5 } $$ is

If $$A = (3 - 4i)$$ and $$B = (9 + ki)$$, where $$k$$ is a constant. 

The simplest form of $$\sqrt {-18} \times \sqrt {-50}$$ is

Simplify $$(2+8i)(1-4i)-(3-2i)(6+4i)$$