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- Question 1, Exercise 1.1
- Question 2 & 3, Exercise 1.1
- Question 4, Exercise 1.1
- Question 5, Exercise 1.1
- Question 6, Exercise 1.1
- Question 7, Exercise 1.1
- Question 8, Exercise 1.1
- Question 9 & 10, Exercise 1.1
- Question 11, Exercise 1.1
- Question 1, Exercise 1.2
- Question 2, Exercise 1.2
- Question 3 & 4, Exercise 1.2
- Question 5, Exercise 1.2
- Question 6, Exercise 1.2
- Question 7, Exercise 1.2
- Question 8, Exercise 1.2
- Question 9, Exercise 1.2
- Question 1, Exercise 1.3
- Question 2, Exercise 1.3
- Question 3 & 4, Exercise 1.3
- Question 5, Exercise 1.3
- Question 6, Exercise 1.3
- Question 1, Review Exercise 1
- Question 2 & 3, Review Exercise 1
- Question 4 & 5, Review Exercise 1
- Question 6, 7 & 8, Review Exercise 1
Fulltext results:
- Question 1, Exercise 1.3
- plex 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{align} Multiply $2$ by (i), we get\\... s_{-}3w&=\mathop-\limits_{+}5i&\mathop+\limits_{-}11 \\ \hline 0&-11w&=11i &-11\\ \end{array} \] \begin{align}-11w&=11i-11\\ \implies w&=\dfrac{11-11i}{
- Question 1, Exercise 1.1
- 1 \right)\cdot i\cdot{{\left( {{i}^{2}} \right)}^{11}}\\ &=-i\cdot{{\left( {{i}^{2}} \right)}^{11}}\\ &=-i\cdot{{\left( -1 \right)}^{11}}\\ &=-i\cdot\left( -1 \right)\\ &=i\end{align} GOOD =====Quest... i}^{-23}} =\frac{1}{i{{\left( {{i}^{2}} \right)}^{11}}}\\ &=\frac{1}{i{{\left( -1 \right)}^{11}}} =\fr
- Question 5, Exercise 1.1
- uestion 5(i)===== Multiply the complex number $8i+11,-7+5i$. ====Solution==== \begin{align}&(8i+11)\times (-7+5i)\\ &=\left( 11+8i \right)\times \left( -7+5i \right)\\ &=\left( -77+40{{i}^{2}} \right)... =\left( -77-40 \right)+\left( 55-56 \right)i\\ &=-117-i\end{align} =====Question 5(ii)===== Multiply
- Question 11, Exercise 1.1
- ====== Question 11, Exercise 1.1 ====== Solutions of Question 11 of Exercise 1.1 of Unit 01: Complex Numbers. This is un... (KPTB or KPTBB) Peshawar, Pakistan. =====Question 11(i)===== Let ${{z}_{1}}=2-i$, ${{z}_{2}}=-2+i$. ... }_{1}}}} \right)=\dfrac{-2}{5}.$$ =====Question 11(ii)===== Let $z_1=2-i$. Find ${\rm Im}\left( \dfr
- Question 5, Exercise 1.3
- frac{-1\pm \sqrt{1-12}}{2}\\ &=\dfrac{-1\pm \sqrt{11}}{2}i\end{align} Thus the solutions of the given equation are $-\dfrac{1}{2}\pm\dfrac{\sqrt{11}}{2}i$. =====Question 5(ii)===== Find the soluti... === <text align="left"><btn type="primary">[[math-11-kpk:sol:unit01:ex1-3-p3 |< Question 3 & 4]]</btn>... t> <text align="right"><btn type="success">[[math-11-kpk:sol:unit01:ex1-3-p5|Question 6 >]]</btn></tex
- Question 9 & 10, Exercise 1.1
- === <text align="left"><btn type="primary">[[math-11-kpk:sol:unit01:ex1-1-p7|< Question 8]]</btn></tex... t> <text align="right"><btn type="success">[[math-11-kpk:sol:unit01:ex1-1-p9|Question 11 >]]</btn></text>
- Question 2 & 3, Exercise 1.1
- ====Question 2===== Prove that ${{i}^{107}}+{{i}^{112}}+{{i}^{122}}+{{i}^{153}}=0$. GOOD ====Solution==== \begin{align}L.H.S.&={{i}^{107}}+{{i}^{112}}+{{i}^{122}}+{{i}^{153}}\\ &=i\cdot i^{106}+i^{112}+i^{122}+i\cdot i^{152}\\ &=i.{{\left( {{i}^{2}}... = |<text align="left"><btn type="primary">[[math-11-kpk:sol:unit01:ex1-1-p1|< Question 1]]</btn></tex
- Question 4, Exercise 1.1
- n} <text align="left"><btn type="primary">[[math-11-kpk:sol:unit01:ex1-1-p2 |< Question 2 & 3]]</btn>... t> <text align="right"><btn type="success">[[math-11-kpk:sol:unit01:ex1-1-p4|Question 5 >]]</btn></tex
- Question 6, Exercise 1.1
- n} <text align="left"><btn type="primary">[[math-11-kpk:sol:unit01:ex1-1-p4|< Question 5]]</btn></tex... t> <text align="right"><btn type="success">[[math-11-kpk:sol:unit01:ex1-1-p6|Question 7 >]]</btn></tex
- Question 7, Exercise 1.1
- === <text align="left"><btn type="primary">[[math-11-kpk:sol:unit01:ex1-1-p5|< Question 6]]</btn></tex... t> <text align="right"><btn type="success">[[math-11-kpk:sol:unit01:ex1-1-p7|Question 8 >]]</btn></tex
- Question 8, Exercise 1.1
- -7i}{4-7i}\\ &=\dfrac{\left( 64-14 \right)-\left( 112+8 \right)i}{16+49}\\ &=\dfrac{50-120i}{65}\\ &=\... === <text align="left"><btn type="primary">[[math-11-kpk:sol:unit01:ex1-1-p6|< Question 7]]</btn></tex... t> <text align="right"><btn type="success">[[math-11-kpk:sol:unit01:ex1-1-p8|Question 9 & 10 >]]</btn
- Question 2, Exercise 1.2
- === <text align="left"><btn type="primary">[[math-11-kpk:sol:unit01:ex1-2-p1|< Question 1]]</btn></tex... t> <text align="right"><btn type="success">[[math-11-kpk:sol:unit01:ex1-2-p3|Question 3, 4 >]]</btn><
- Question 3 & 4, Exercise 1.2
- === <text align="left"><btn type="primary">[[math-11-kpk:sol:unit01:ex1-2-p2|< Question 2]]</btn></tex... t> <text align="right"><btn type="success">[[math-11-kpk:sol:unit01:ex1-2-p4|Question 5 >]]</btn></te
- Question 5, Exercise 1.2
- === <text align="left"><btn type="primary">[[math-11-kpk:sol:unit01:ex1-2-p3|< Question 3 & 4]]</btn><... t> <text align="right"><btn type="success">[[math-11-kpk:sol:unit01:ex1-2-p5|Question 6 >]]</btn></te
- Question 6, Exercise 1.2
- === <text align="left"><btn type="primary">[[math-11-kpk:sol:unit01:ex1-2-p4|< Question 5]]</btn></tex... t> <text align="right"><btn type="success">[[math-11-kpk:sol:unit01:ex1-2-p6|Question 7 >]]</btn></te