Complex Numbers Class 11 Important modulus and conjugate Questions with Solutions

Complex Numbers is one of the most important chapters in Class 11 Mathematics, forming the foundation for competitive exams like JEE, NDA, IMU CET. To help students strengthen their concepts and improve problem-solving skills, we have compiled Complex Numbers Class 11 Important Questions with Solutions. These carefully selected questions cover all key concepts, including imaginary numbers, algebra of complex numbers, modulus, argument, conjugate, polar form, and quadratic equations, along with step-by-step solutions for better understanding and effective exam preparation.

Problem
Find the modulus of the complex number : ${\displaystyle \frac{1}{2-3i}}$

Solution. Let $z={\displaystyle \frac{1}{2-3i}}$

$z={\displaystyle \frac{2+3i}{(2-3i)(2+3i)}}={\displaystyle \frac{2+3i}{4-9i^2}}={\displaystyle \frac{2+3i}{4+9}}={\displaystyle \frac{1}{13}}(2+3i).$

Hence, $|z|={\displaystyle \frac{1}{13}}\sqrt{4+9}={\displaystyle \frac{1}{13}}\sqrt{13}={\displaystyle \frac{1}{\sqrt{13}}}.$

Understand related topics like Modulus of Complex Number Properties Explain With Proof

---

Problem
If $z_1=3+4i$ and $z_2=12-5i$, verify:
(i) $|-z_1|=|z_1|$
(ii) $|z_{1}+z_{2}| < |z_{1}|+|z_{2}|$
(iii) $|z_1{z}_2|=|z_1||z_2|$

Solution. We have: $z_1=3+4i$ and $z_2=12-5i$.

(i) $-z_1=-3-4i.$

Also,

$|-z_1|=\sqrt{(-3{)}^2+(-4{)}^2}=\sqrt{9+16}=\sqrt{25}=5.$

$|z_1|=\sqrt{3^2+4^2}=\sqrt{9+16}=\sqrt{25}=5.$

Hence, $|-z_1|=|z_1|.$

(ii) $z_1+z_2=(3+4i)+(12-5i)=15-i$

$|z_1+z_2|=\sqrt{(15{)}^2+(-1{)}^2}=\sqrt{225+1}=\sqrt{226}.$

Also, $|z_1|+|z_2|=5+\sqrt{144+25}=5+13=18.$

Hence, $|z_{1}+z_{2}| < |z_{1}|+|z_{2}|$ because √226 < 18

(iii) $z_1{z}_2=(3+4i)(12-5i)$

$z_1{z}_2=36-15i+48i-20i^2=(36+20)+33i=56+33i.$

Also,

$|z_1{z}_2|=\sqrt{(56{)}^2+(33{)}^2}=\sqrt{3136+1089}=\sqrt{4225}=65$

and $|z_1|=\sqrt{9+16}=5$, $|z_2|=\sqrt{144+25}=13.$

Hence, $|z_1{z}_2|=|z_1||z_2|[\text{since }65=(5)(13)].$

Continue learning with Complex Numbers: Addition, Subtraction, Multiplication of Two Complex Numbers With Solved Examples

---

Problem
If $z_1,z_2$ are complex numbers such that ${\displaystyle \frac{2z_1}{3z_2}}$ is a purely imaginary number, then find $\left|{\displaystyle \frac{z_1-z_2}{z_1+z_2}}\right|.$

Solution. Since ${\displaystyle \frac{2z_1}{3z_2}}$ is purely imaginary,

${\displaystyle \frac{2z_1}{3z_2}}=ki,\text{ where }k\in ℝ\text{[Given]}$

$\Rightarrow {\displaystyle \frac{z_1}{z_2}}={\displaystyle \frac{3k}{2}}i\dots (1)$

Now, $\left|{\displaystyle \frac{z_1-z_2}{z_1+z_2}}\right|=\left|{\displaystyle \frac{{\displaystyle \frac{z_1}{z_2}}-1}{{\displaystyle \frac{z_1}{z_2}}+1}}\right|=\left|{\displaystyle \frac{{\displaystyle \frac{3k}{2}}i-1}{{\displaystyle \frac{3k}{2}}i+1}}\right|\text{[Using (1)]}$

$=\left|{\displaystyle \frac{{\displaystyle \frac{3ki-2}{2}}}{{\displaystyle \frac{3ki+2}{2}}}}\right|=\left|{\displaystyle \frac{-2+3ki}{2+3ki}}\right|$

$={\displaystyle \frac{|-2+3ki|}{|2+3ki|}}={\displaystyle \frac{\sqrt{(-2{)}^2+(3k{)}^2}}{\sqrt{(2{)}^2+(3k{)}^2}}}$

$={\displaystyle \frac{\sqrt{4+9k^2}}{\sqrt{4+9k^2}}}=1.$

---

Problem
If $z=x+iy$ and $w={\displaystyle \frac{1-iz}{z-i}},$ show that: $|w|=1\Rightarrow z$ is purely real.

Solution. Since $|w|=1\Rightarrow \left|{\displaystyle \frac{1-iz}{z-i}}\right|=1$

$\Rightarrow {\displaystyle \frac{|1-iz|}{|z-i|}}=1\Rightarrow |1-iz|=|z-i|$

$\Rightarrow |1-i(x+iy)|=|x+iy-i|$

$\Rightarrow |(1+y)-ix|=|x+i(y-1)|$

$\Rightarrow \sqrt{(1+y{)}^2+x^2}=\sqrt{x^2+(y-1{)}^2}$

Squaring, $(1+y{)}^2+x^2=x^2+(y-1{)}^2$

$\Rightarrow 1+y^2+2y+x^2=x^2+y^2-2y+1$

$\Rightarrow 4y=0\Rightarrow y=0.$

Hence, $z=x+iy=x+i(0)=x,$ which is purely real.

---

Problem
If $(1+i)(1+2i)(1+3i)\dots (1+ni)=(x+iy)$, show that:
$2\cdot 5\cdot 10\dots (1+n^2)=x^2+y^2.$

Solution. Given, $(1+i)(1+2i)(1+3i)\dots (1+ni)=x+iy$

$\Rightarrow |(1+i)(1+2i)(1+3i)\dots (1+ni)|=|x+iy|$

$\Rightarrow |1+i|\cdot |1+2i|\cdot |1+3i|\dots |1+ni|=|x+iy|$

$[\because |z_1{z}_2\dots z_n|=|z_1||z_2|\dots |z_n|]$

$\Rightarrow \sqrt{1+1}\cdot \sqrt{1+4}\cdot \sqrt{1+9}\dots \sqrt{1+n^2}=\sqrt{x^2+y^2}.$

Squaring, $2\cdot 5\cdot 10\dots (1+n^2)=x^2+y^2,$ which is true.

---

Problem
Reduce $({\displaystyle \frac{1}{1+2i}}+{\displaystyle \frac{3}{1-i}})\left({\displaystyle \frac{3-2i}{1+3i}}\right)$ to the form $(a+ib)$.

Solution. We have

$({\displaystyle \frac{1}{1+2i}}+{\displaystyle \frac{3}{1-i}})\left({\displaystyle \frac{3-2i}{1+3i}}\right)$

$=\left\{{\displaystyle \frac{(1-i)+(3+6i)}{(1+2i)(1-i)}}\right\}\left({\displaystyle \frac{3-2i}{1+3i}}\right)={\displaystyle \frac{(4+5i)}{(3+i)}}\times {\displaystyle \frac{(3-2i)}{(1+3i)}}$

$={\displaystyle \frac{(4+5i)(3-2i)}{(3+i)(1+3i)}}={\displaystyle \frac{(12+10)+(15-8)i}{(3-3)+(9i+i)}}={\displaystyle \frac{(22+7i)}{10i}}\times {\displaystyle \frac{i}{i}}$

$={\displaystyle \frac{(7i^2+22i)}{10i^2}}={\displaystyle \frac{-7+22i}{-10}}=({\displaystyle \frac{7}{10}}-{\displaystyle \frac{22}{10}}i)=({\displaystyle \frac{7}{10}}-{\displaystyle \frac{11}{5}}i)$

---

Problem
Reduce $({\displaystyle \frac{1+i}{1-i}}-{\displaystyle \frac{1-i}{1+i}})$ to the form $(a+ib)$ and hence find its modulus.

Solution. Let $z=({\displaystyle \frac{1+i}{1-i}}-{\displaystyle \frac{1-i}{1+i}})={\displaystyle \frac{(1+i{)}^2-(1-i{)}^2}{(1-i)(1+i)}}={\displaystyle \frac{(4\times 1\times i)}{(1-i^2)}}={\displaystyle \frac{4i}{2}}=2i.$

Thus, $z=0+2i\Rightarrow |z|=\sqrt{0^2+2^2}=\sqrt{4}=2$.

Hence, $z=0+2i$ and $|z|=2$.

---

Problem
If $\sqrt{{\displaystyle \frac{1+i}{1-i}}}=(a+ib)$ then show that $(a^2+b^2)=1$.

Solution. We have

$(a+ib)=\sqrt{{\displaystyle \frac{1+i}{1-i}}}={\displaystyle \frac{\sqrt{1+i}}{\sqrt{1-i}}}\times {\displaystyle \frac{\sqrt{1+i}}{\sqrt{1+i}}}={\displaystyle \frac{(1+i)}{\sqrt{1-i^2}}}$

$={\displaystyle \frac{(1+i)}{\sqrt{2}}}=({\displaystyle \frac{1}{\sqrt{2}}}+{\displaystyle \frac{1}{\sqrt{2}}}i)$

$\Rightarrow |a+ib{|}^2={\left({\displaystyle \frac{1}{\sqrt{2}}}\right)}^2+{\left({\displaystyle \frac{1}{\sqrt{2}}}\right)}^2=({\displaystyle \frac{1}{2}}+{\displaystyle \frac{1}{2}})=1.$

$\Rightarrow (a^2+b^2)=1.$

Hence, $(a^2+b^2)=1$.

---

Problem
Find the least positive integer m for which ${\left({\displaystyle \frac{1+i}{1-i}}\right)}^m=1$.

Solution. We have

$\left({\displaystyle \frac{1+i}{1-i}}\right)={\displaystyle \frac{(1+i)}{(1-i)}}\times {\displaystyle \frac{(1+i)}{(1+i)}}={\displaystyle \frac{(1+i{)}^2}{(1-i^2)}}={\displaystyle \frac{(1+2i+i^2)}{2}}={\displaystyle \frac{2i}{2}}=i.$

${\left({\displaystyle \frac{1+i}{1-i}}\right)}^m=1\Rightarrow i^m=1$

And, we know that $4$ is the least positive integer such that $i^4=1$ and therefore, m = 4.

---

Problem
Separate $\left({\displaystyle \frac{3+\sqrt{-1}}{2-\sqrt{-1}}}\right)$ into real and imaginary parts and hence find its modulus.

Solution. Let $z=\left({\displaystyle \frac{3+\sqrt{-1}}{2-\sqrt{-1}}}\right)=\left({\displaystyle \frac{3+i}{2-i}}\right)={\displaystyle \frac{(3+i)}{(2-i)}}={\displaystyle \frac{(3+i)}{(2-i)}}\times {\displaystyle \frac{(2+i)}{(2+i)}}$

$={\displaystyle \frac{(3+i)(2+i)}{(2-i)(2+i)}}={\displaystyle \frac{(6+i^2)+5i}{(4-i^2)}}={\displaystyle \frac{5+5i}{\{4-(-1)\}}}={\displaystyle \frac{5(1+i)}{5}}=(1+i).$

$|z|=\sqrt{1^2+1^2}=\sqrt{2}$

Hexnce, $z=(1+i)$ and $|z|=\sqrt{2}$.

---

Problem
Reduce $\left\{{\displaystyle \frac{\sqrt{5+12i}+\sqrt{5-12i}}{\sqrt{5+12i}-\sqrt{5-12i}}}\right\}$ to the form $(a+ib)$ and hence find its conjugate.

Solution. We have

$z={\displaystyle \frac{\sqrt{5+12i}+\sqrt{5-12i}}{\sqrt{5+12i}-\sqrt{5-12i}}}$

$={\displaystyle \frac{(\sqrt{5+12i}+\sqrt{5-12i})}{(\sqrt{5+12i}-\sqrt{5-12i})}}\times {\displaystyle \frac{(\sqrt{5+12i}+\sqrt{5-12i})}{(\sqrt{5+12i}+\sqrt{5-12i})}}$

$={\displaystyle \frac{(\sqrt{5+12i}+\sqrt{5-12i}{)}^2}{(5+12i)-(5-12i)}}$

$={\displaystyle \frac{(5+12i)+(5-12i)+2\sqrt{(5{)}^2-(12i{)}^2}}{24i}}$

$={\displaystyle \frac{10+2\sqrt{25+144}}{24i}}={\displaystyle \frac{10+2\sqrt{169}}{24i}}={\displaystyle \frac{(10+2\times 13)}{24i}}$

$={\displaystyle \frac{36}{24i}}={\displaystyle \frac{3}{2i}}={\displaystyle \frac{3}{2i}}\times {\displaystyle \frac{i}{i}}={\displaystyle \frac{3i}{2i^2}}=-{\displaystyle \frac{3}{2}}i.$

Hence, $z=(0-{\displaystyle \frac{3}{2}}i)$ and $\overline{z}=\overline{(0-{\displaystyle \frac{3}{2}}i)}=(0+{\displaystyle \frac{3}{2}}i)$.

---

Problem
If $(x+iy{)}^{1/3}=(a+ib)$ then prove that:
1. ${\displaystyle \frac{x}{a}}+{\displaystyle \frac{y}{b}}=4(a^2-b^2)$
2. ${\displaystyle \frac{x}{a}}-{\displaystyle \frac{y}{b}}=-2(a^2+b^2)$

Solution. We have

$(x+iy{)}^{1/3}=(a+ib)$

$\Rightarrow (x+iy)=(a+ib{)}^3\text{[on cubing both sides]}$

$\Rightarrow (x+iy)=a^3+i^3{b}^3+3iab(a+ib)$

$=a^3-i{b}^3+3a^{2}bi-3a{b}^2=(a^3-3a{b}^2)+i(3a^{2}b-b^3)$

$\Rightarrow x=a^3-3a{b}^2\text{ and }y=3a^{2}b-b^3$

[on equating real and imaginary parts separately]

$\Rightarrow {\displaystyle \frac{x}{a}}=(a^2-3b^2)\text{ and }{\displaystyle \frac{y}{b}}=(3a^2-b^2)$

Hence,

1. $({\displaystyle \frac{x}{a}}+{\displaystyle \frac{y}{b}})=4(a^2-b^2)$

2. $({\displaystyle \frac{x}{a}}-{\displaystyle \frac{y}{b}})=-2(a^2+b^2)$.

---

Problem
Let $z$ be a complex number such that $z\ne 1$ and $|z|=1$. Then, prove that $\left({\displaystyle \frac{z-1}{z+1}}\right)$ is purely imaginary. What will be your conclusion when $z=1$?

Solution. Let $z=(x+iy)$ be the given complex number such that $z\ne 1$ and $|z|=1$. Then,

$|z|=1\Rightarrow |z{|}^2=1\Rightarrow x^2+y^2=1\Rightarrow x^2+y^2-1=0.\dots \text{(i)}$

$\left({\displaystyle \frac{z-1}{z+1}}\right)={\displaystyle \frac{x+iy-1}{x+iy+1}}={\displaystyle \frac{(x-1)+iy}{(x+1)+iy}}\times {\displaystyle \frac{(x+1)-iy}{(x+1)-iy}}$

$={\displaystyle \frac{\{(x-1)+iy\}\{(x+1)-iy\}}{\{(x+1{)}^2+y^2\}}}$

$={\displaystyle \frac{(x^2-1+y^2)+i\{(x+1)y-(x-1)y\}}{\{(x+1{)}^2+y^2\}}}$

$={\displaystyle \frac{(x^2+y^2-1)+(2y)i}{\{(x+1{)}^2+y^2\}}}$

$=\left\{{\displaystyle \frac{2yi}{(x+1{)}^2+y^2}}\right\}\text{[using (i)]},$

which is purely imaginary.

Particular Case : When $z=1$.

In this case, $z=1\Rightarrow x+iy=1$

$\Rightarrow (x-1)+iy=0\Rightarrow x-1=0,y=0\Rightarrow x=1,y=0.$

${\displaystyle \frac{z-1}{z+1}}={\displaystyle \frac{(x+iy)-1}{(x+iy)+1}}={\displaystyle \frac{(x-1)+iy}{(x+1)+iy}}={\displaystyle \frac{(1-1)+i\times 0}{(1+1)+i\times 0}}={\displaystyle \frac{0}{2}}=0.$

Thus, $z=1\Rightarrow {\displaystyle \frac{z-1}{z+1}}$ is purely real.

---

Problem
Find the real value of $\theta$ for which $\left({\displaystyle \frac{3+2i\sin \theta }{1-2i\sin \theta }}\right)$ is purely real.

Solution. We have

$\left({\displaystyle \frac{3+2i\sin \theta }{1-2i\sin \theta }}\right)={\displaystyle \frac{(3+2i\sin \theta )}{(1-2i\sin \theta )}}\times {\displaystyle \frac{(1+2i\sin \theta )}{(1+2i\sin \theta )}}$

$={\displaystyle \frac{(3+2i\sin \theta )(1+2i\sin \theta )}{(1-4i^2{\sin }^2\theta )}}$

$={\displaystyle \frac{(3-4{\sin }^2\theta )+i(6\sin \theta +2\sin \theta )}{(1+4{\sin }^2\theta )}}$

$={\displaystyle \frac{(3-4{\sin }^2\theta )+i(8\sin \theta )}{(1+4{\sin }^2\theta )}}$

Now, $\left({\displaystyle \frac{3+2i\sin \theta }{1-2i\sin \theta }}\right)$ will be purely real only when

${\displaystyle \frac{8\sin \theta }{(1+4{\sin }^2\theta )}}=0.$

This happens only when $8\sin \theta =0\iff \sin \theta =0\iff \theta =n\pi ,n\in \mathbb{Z}.$

Hence, the required value of $\theta$ is $n\pi$, where $n\in \mathbb{Z}$.

---

Problem
Show that $|1-i{|}^x=2^x$ has no nonzero integral solution.

Solution.

$|1-i{|}^x=2^x\Rightarrow (\sqrt{2}{)}^x=2^x[\because |1-i|=\sqrt{1^2+(-1{)}^2}=\sqrt{2}]$

$\Rightarrow 2^{x/2}=2^x\Rightarrow {\displaystyle \frac{2^x}{2^{x/2}}}=1$

$\Rightarrow 2^{x/2}=1=2^0\Rightarrow {\displaystyle \frac{x}{2}}=0\Rightarrow x=0$

Thus, $x=0$ is the only solution of the given equation.

Hence, the given equation has no nonzero integral solution.

---

Problem
Solve for $x$ and $y$:
1. $(3x-7)+2iy=-5y+(5+x)i$
2. $(x+iy)(2-3i)=4+i$

Solution.

(i) The given equation is

$(3x-7)+2iy=-5y+(5+x)i.$

Equating the real parts and the imaginary parts of the given equation separately, we get

$3x-7=-5y\text{ and }2y=5+x$

$\Rightarrow 3x+5y=7\text{ and }x-2y=-5$

$\Rightarrow x=-1\text{ and }y=2\text{[on solving }3x+5y=7,x-2y=-5]$

Hence, $x=-1$ and $y=2$.

(ii) The given equation is

$(x+iy)(2-3i)=4+i$

$\Rightarrow (2x+3y)+i(2y-3x)=4+i$

$\Rightarrow 2x+3y=4\text{ and }2y-3x=1$

[on equating real and imaginary parts separately]

$\Rightarrow 2x+3y=4\text{ and }-3x+2y=1$

$\Rightarrow x={\displaystyle \frac{5}{13}}\text{ and }y={\displaystyle \frac{14}{13}}\text{[on solving }2x+3y=4,-3x+2y=1].$

Hence, $x={\displaystyle \frac{5}{13}}$ and $y={\displaystyle \frac{14}{13}}$ is the required solution.

---

Problem
Find the real values of x and y for which $({\displaystyle \frac{x-1}{3+i}}+{\displaystyle \frac{y-1}{3-i}})=i.$

Solution. We have

${\displaystyle \frac{x-1}{3+i}}+{\displaystyle \frac{y-1}{3-i}}=i$

$\Rightarrow {\displaystyle \frac{(x-1)}{(3+i)}}\times {\displaystyle \frac{(3-i)}{(3-i)}}+{\displaystyle \frac{(y-1)}{(3-i)}}\times {\displaystyle \frac{(3+i)}{(3+i)}}=i$

$\Rightarrow {\displaystyle \frac{(x-1)(3-i)}{(9-i^2)}}+{\displaystyle \frac{(y-1)(3+i)}{(9-i^2)}}=i$

$\Rightarrow {\displaystyle \frac{3(x-1)-i(x-1)}{10}}+{\displaystyle \frac{3(y-1)+i(y-1)}{10}}=i$

$\Rightarrow 3(x-1)-i(x-1)+3(y-1)+i(y-1)=10i$

$\Rightarrow (3x-3+3y-3)+i(y-1-x+1)=10i$

$\Rightarrow (3x+3y-6)+i(y-x)=10i$

$\Rightarrow 3(x+y-2)=0\text{ and }y-x=10$

[equating real parts and the imaginary parts separately]

$\Rightarrow x+y-2=0\text{ and }y-x=10$

$\Rightarrow x+y=2\text{ and }-x+y=10$

$\Rightarrow x=-4\text{ and }y=6$

[on solving $x+y=2$, $-x+y=10$].

Hence, $x=-4$ and $y=6$ are the required values.

---

Problem
Find the complex number z for which $|z+1|=z+2(1+i)$.

Solution. Let the required complex number be $z=(x+iy)$. Then,

$|z+1|=z+2(1+i)$

$\Rightarrow |(x+iy)+1|=(x+iy)+2(1+i)$

$\Rightarrow \sqrt{(x+1{)}^2+y^2}=(x+2)+i(y+2)$

$\Rightarrow \sqrt{(x+1{)}^2+y^2}=(x+2)\text{ and }y+2=0$

[equating real parts and imaginary parts separately]

$\Rightarrow y=-2\text{ and }\sqrt{(x+1{)}^2+(-2{)}^2}=(x+2)$

$\Rightarrow y=-2\text{ and }\sqrt{x^2+2x+5}=(x+2)$

$\Rightarrow y=-2\text{ and }(x^2+2x+5)=(x+2{)}^2$

$\Rightarrow x^2+2x+5=x^2+4x+4\text{ and }y=-2$

$\Rightarrow 2x=1\text{ and }y=-2\Rightarrow x={\displaystyle \frac{1}{2}}\text{ and }y=-2.$

Hence, the required complex number is $z=({\displaystyle \frac{1}{2}}-2i)$ .

---

Problem
Solve the equation $|z|+z=(2+i)$ for complex value of z.

Solution. Let $z=(x+iy)$. Then,

$|z|+z=2+i$

$\Rightarrow |x+iy|+(x+iy)=2+i$

$\Rightarrow \{\sqrt{x^2+y^2}+x\}+iy=(2+i)$

$\Rightarrow \sqrt{x^2+y^2}+x=2\text{ and }y=1$

[equating real parts and imaginary parts separately]

$\Rightarrow y=1\text{ and }\sqrt{x^2+1^2}+x=2$

$\Rightarrow y=1\text{ and }\sqrt{x^2+1}=(2-x)$

$\Rightarrow y=1\text{ and }{x}^2+1=(2-x{)}^2$

$\Rightarrow y=1\text{ and }{x}^2+1=x^2-4x+4$

$\Rightarrow 4x=3\text{ and }y=1\Rightarrow x={\displaystyle \frac{3}{4}}\text{ and }y=1.$

Hence, $z=({\displaystyle \frac{3}{4}}+i)$ is the desired solution.

---

Problem
Solve the equation $z+\sqrt{2}|z+1|+i=0$ for complex value of z.

Solution. Let the required complex number be $z=x+iy$ Then,

$z+\sqrt{2}|z+1|+i=0$

$\Rightarrow (x+iy)+\sqrt{2}|(x+iy)+1|+i=0$

$\Rightarrow x+\sqrt{2}\{\sqrt{(x+1{)}^2+y^2}\}+(y+1)i=0$

$\Rightarrow x+\sqrt{2}\{\sqrt{(x+1{)}^2+y^2}\}=0\text{ and }y+1=0$

[equating real parts and imaginary parts separately]

$\Rightarrow y=-1\text{ and }\sqrt{2}\{\sqrt{(x+1{)}^2+(-1{)}^2}\}=(-x)$

$\Rightarrow y=-1\text{ and }\sqrt{2}\cdot \sqrt{x^2+2x+2}=(-x)$

$\Rightarrow y=-1\text{ and }2(x^2+2x+2)=x^2$

[squaring on both sides]

$\Rightarrow y=-1\text{ and }{x}^2+4x+4=0$

$\Rightarrow (x+2{)}^2=0\text{ and }y=-1$

$\Rightarrow x+2=0\text{ and }y=-1\Rightarrow x=-2\text{ and }y=-1.$

Hence, the required complex number is $z=(-2-i)$.

---

Problem
If $z=3+2i$, prove that $z^2-6z+13=0$ and hence deduce that $3z^3-13z^2+9z+65=0.$

Solution.

$z=3+2i\Rightarrow z-3=2i\Rightarrow (z-3{)}^2=4i^2$

$\Rightarrow z^2-6z+13=0.$

Thus, $z^2-6z+13=0$ … (i)

Now, $3z^3-13z^2+9z+65$

$=3z(z^2-6z+13)+5(z^2-6z+13)$

$=(3z\times 0)+(5\times 0)=0$

[using (i)].

Hence, $3z^3-13z^2+9z+65=0.$

---

Problem
If $z=-5+3i$, find the value of $(z^4+9z^3+26z^2-14z+8)$ .

Solution. We have

$z=-5+3i\Rightarrow (z+5)=3i$

$\Rightarrow (z+5{)}^2=9i^2\Rightarrow z^2+10z+34=0.$ … (i)

Now, $z^4+9z^3+26z^2-14z+8$

$=z^2(z^2+10z+34)-z(z^2+10z+34)+2(z^2+10z+34)-60$

$=(z^2\times 0)-(z\times 0)+(2\times 0)-60=-60$

[using (i)].

Hence, the value of $z^4+9z^3+26z^2-14z+8$ is -60.

---

Problem
If $z=2+i$, prove that $z^3+3z^2-9z+8=(1+14i)$.

Solution. We have

$z=2+i\Rightarrow z-2=i$

$\Rightarrow (z-2{)}^2=i^2\Rightarrow z^2-4z+5=0.$ … (i)

Now, $z^3+3z^2-9z+8$

$=z(z^2-4z+5)+7(z^2-4z+5)+14z-27$

$=(z\times 0)+(7\times 0)+14z-27=(14z-27)$

[using (i)]

$=14(2+i)-27=(1+14i).$

Hence, $z^3+3z^2-9z+8=(1+14i)$.

---