Complex numbers and quadratic equations is a segment of maths that deals with crucial theorems and concepts along with various formulae. It comprises of linear and quadratic equations along with roots related to the complex number's set (known as complex roots)..

**Q1.**If z

_{1},z

_{2}and z

_{3}are complex numbers such that |z

_{1}|=|z

_{2}|=|z

_{3}|=|(1/z

_{1})+(1/z

_{2})+(1/z

_{2})|=1, then |z

_{1}+z

_{2}+z

_{3}| is

**Q2.**If z

_{1}z

_{2}be complex numbers such that z

_{1}≠ z

_{2}and |z

_{1}|= |z

_{2}|. If z

_{1}has positive real part and z

_{2}has negative imaginary part, then[(z

_{1}+z

_{2})/(z

_{1}-z

_{2})] may be

**Q3.**If the expression [mx-1+(1/x)] is non-negative for all positive real x, then the minimum value of m must be

Solution

(c) We know that ax

If a > 0 and b

mx -1 + 1/x ≥ 0 ⇒ (mx

⇒mx

Now, mx

⇒ m > 0 and m ≥ 1/4

Thus, the minimum value of m is ¼

(c) We know that ax

^{2}+ bx + c ≥ 0 , ∀ x ∈ R,If a > 0 and b

^{2}- 4ac ≤ 0. So,mx -1 + 1/x ≥ 0 ⇒ (mx

^{2}- x + 1)/x ≥ 0⇒mx

^{2}- x + 1 ≥ 0 as x > 0Now, mx

^{2}- x + 1 ≥ 0 if m > 0 and 1-4m ≤ 0⇒ m > 0 and m ≥ 1/4

Thus, the minimum value of m is ¼

**Q4.**If for complex numbers z

_{1}z

_{2}, arg(z

_{1})-arg(z

_{2})=0, then |z

_{1}-z

_{2}| is equal to

**Q6.**For positive integers n

_{1},n

_{2}the value of the expression (1+i)

^{(n1 )}+ (1 + i

^{3})

^{(n1 )}+ (1 + i

^{5})

^{(n2 )}+ (1 + i

^{7})

^{(n2 )}, where i=√(-1) is a real number if and only if

**Q7.**If |z

_{2}+iz

_{1}|=|z

_{1}|+|z

_{2}| and |z

_{1}|=3 and |z

_{2}|=4, then area of ∆ABC, if affixes of A,B and C are z

_{1},z

_{2}and [(z

_{2}-iz

_{1})/(1-i)] respectively, is

**Q8.**x

^{2}-xy+y

^{2}-4x-4y+16=0 represents

Solution

(a)

Given equation is

x

Since x is real, so,

D≥0

⇒(y+4)

⇒ -3y

⇒

⇒(y-4)

⇒y-4=0

⇒y=4

Since the equation is symmetric in x and y, therefore x=4 only

(a)

Given equation is

x

^{2}-(y+4)x+y^{2}-4y+16=0Since x is real, so,

D≥0

⇒(y+4)

^{2}-4(y^{2}-4y+16)≥0⇒ -3y

^{2}+24y-48≥0⇒

^{^}2-8y+16≤0⇒(y-4)

^{2}≤0⇒y-4=0

⇒y=4

Since the equation is symmetric in x and y, therefore x=4 only

**Q9.**Sum of common roots of the equations z

^{3}+2z

^{2}+2z+1=0 and z

^{1985}+z

^{100}+1=0 is

Solution

(a)

We have,

z

⇒(z

⇒(z+1)(z

⇒z=-1,Ï‰,Ï‰

Since z=-1 does not satisfy z

(a)

We have,

z

^{3}+2z^{2}+2z+1=0⇒(z

^{3}+1)+2z(z+1)=0⇒(z+1)(z

^{2}+z+1)=0⇒z=-1,Ï‰,Ï‰

^{2}Since z=-1 does not satisfy z

^{1985}+z^{100}+1=0 while z=Ï‰,Ï‰^{2}satisfy it, hence sum is Ï‰+Ï‰^{2}=-1**Q10.**If the roots of the equation ax

^{2}-bx+c=0 are Î±,Î² then the roots of the equation b

^{2}cx

^{2}-ab

^{2}x+a

^{3}=0 are