Search
You can find the results of your search below.
Fulltext results:
- Question 1, Exercise 1.3
- Question 1 of Exercise 1.3 of Unit 01: Complex Numbers. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Pesha... aneous linear equation with complex coefficient. \begin{align}&z-4w=3i\\ &2z+3w=11-5i\end{align} ====Solution==== Given that \begin{align}z-4w&=3i …(i)\\ 2z+3w&=11-5i …(ii)\end{
- Question 5, Exercise 1.2
- Question 5 of Exercise 1.2 of Unit 01: Complex Numbers. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Pesha... 1}}}=2-4i$ and $\overline{{{z}_{2}}}=1+3i$. Now \begin{align}z_1+z_2&=2+4i+1-3i\\ &=3+i \end{align} Now \begin{align}\overline{z_1+z_2}=3-i \ldots (1)\end{al
- Question 2, Exercise 1.2
- Question 2 of Exercise 1.2 of Unit 01: Complex Numbers. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Pesha... , that is, $$(z_1+z_2)+z_3=z_1+(z_2+z_3).$$ Take \begin{align} {{z}_{1}}+{{z}_{2}}&=\left( -1+i \right)+\left( 3-2i \right)\\ &=2-i\end{align} So \begin{align} \left( {{z}_{1}}+{{z}_{2}} \right)+{{z}
- Question 8, Exercise 1.2
- Question 8 of Exercise 1.2 of Unit 01: Complex Numbers. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Pesha... ==== Assume $z=a+ib$, then $\overline{z}=a-ib$. \begin{align}z+\overline{z}&=\left( a+ib \right)+\lef... Assume that $z=a+ib$, then $\overline{z}=a-ib$. \begin{align}z-\overline{z}&=\left( a+ib \right)-\lef
- Question 2, Exercise 1.3
- Question 2 of Exercise 1.3 of Unit 01: Complex Numbers. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Pesha... is a factor of $P(z)$ iff $P(a)=0$. Put $z=-2$ \begin{align} P(-2)&=(-2)^3+6(-2)+20\\ &=-8-12+20\\ &... 6z+20$.\\ By using synthetic division, we have $$\begin{array}{c|cccc} -2 & 1 & 0 & 6 & 20 \\ & \do
- Question 2 & 3, Review Exercise 1
- 2 & 3 of Review Exercise 1 of Unit 01: Complex Numbers. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Pesha... {n+3}}=0$, $\forall n\in N$ \\ ====Solution==== \begin{align}{{i}^{n}}+{{i}^{n+1}}+{{i}^{n+2}}+{{i}^{... =====Question 3(i)===== Express the complex number $\left( 1+3i \right)+\left( 5+7i \right)$ in th
- Question 2 & 3, Exercise 1.1
- tion 2 & 3 of Exercise 1.1 of Unit 01: Complex Numbers. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Pesha... {i}^{122}}+{{i}^{153}}=0$. GOOD ====Solution==== \begin{align}L.H.S.&={{i}^{107}}+{{i}^{112}}+{{i}^{12... GOOD =====Question 3(i)===== Add the complex numbers $3\left( 1+2i \right),-2\left( 1-3i \right)$. =
- Question 6, Exercise 1.2
- Question 6 of Exercise 1.2 of Unit 01: Complex Numbers. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Pesha... ==Question 6(i)===== Show that for all complex numbers ${{z}_{1}}$and ${{z}_{2}}$. Verify that $|{{z}_... rt{a^2+b^2}|$ and $|z_2=\sqrt{c^2+d^2}|$.\\ Now \begin{align} L.H.S.&=|{{z}_{1}}{{z}_{2}}|\\ &=|(a+bi
- Question 6, Exercise 1.3
- Question 6 of Exercise 1.3 of Unit 01: Complex Numbers. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Pesha... ====Solution==== Given: $$z^3=-8.$$ This gives \begin{align} & z^3+2^3=0\\ \implies &(z+2)\left(z^2-2z+4 \right)=0 \\ & \quad \because a^3+b^3=\left( a+b \right)\left( {{a}^{2}}-a
- Question 4, Exercise 1.1
- Question 4 of Exercise 1.1 of Unit 01: Complex Numbers. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Pesha... Question 4(i)===== Subtract the second complex number from first $\left( a,0 \right)\left( 2,-b \right)$. ====Solution==== \begin{align}&\left( a,0 \right)-\left( 2,-b \right)\
- Question 5, Exercise 1.1
- Question 5 of Exercise 1.1 of Unit 01: Complex Numbers. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Pesha... . =====Question 5(i)===== Multiply the complex number $8i+11,-7+5i$. ====Solution==== \begin{align}&(8i+11)\times (-7+5i)\\ &=\left( 11+8i \right)\times \
- Question 7, Exercise 1.1
- Question 7 of Exercise 1.1 of Unit 01: Complex Numbers. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Pesha... === We know that $z_1=1+2i$ and $z_2=2+3i$, then \begin{align} {{z}_{1}}+{{z}_{2}}&=1+2i+2+3i\\ &=1+2+2i+3i\\ &=3+5i \end{align} Now \begin{align} |z_1+z_2|&=\sqrt{3^2+5^2}\\ &=\sqrt{9+2
- Question 7, Exercise 1.2
- Question 7 of Exercise 1.2 of Unit 01: Complex Numbers. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Pesha... ry parts $\dfrac{2+3i}{5-2i}$. ====Solution==== \begin{align}&\dfrac{2+3i}{5-2i} \\ =&\dfrac{2+3i}{5-... t( 1+2i \right)}^{2}}}{1-3i}$. ====Solution==== \begin{align}&\dfrac{(1+2i)^2}{1-3i}\\ =&\dfrac{1-4+4
- Question 9 & 10, Exercise 1.1
- ion 9 & 10 of Exercise 1.1 of Unit 01: Complex Numbers. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Pesha... right)\left( 2-i \right)}$. ====Solution==== Let \begin{align}z&=\dfrac{\left( 3-2i \right)\left( 2+3i... 2+2+4i-i}\\ &=\dfrac{12+5i}{4+3i}\end{align} Now \begin{align}\bar{z}&=\dfrac{12-5i}{4-3i}\\ &=\dfrac{
- Question 3 & 4, Exercise 1.2
- tion 3 & 4 of Exercise 1.2 of Unit 01: Complex Numbers. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Pesha... ive property w.r.t. addition and multiplicative. \begin{align}{{z}_{1}}\left( {{z}_{2}}+{{z}_{3}} \rig... tive and multiplicative inverse of the complex number $5+2i$. ====Solution==== Given $z=5+2i$. Here $